5.1 Sequences

1. Sequences

Title

Calculus II · Section 5.1

Functions on the whole numbers, and where they are heading

2. What this lesson gives you

Objectives

Chapter 5 is about adding up infinitely many numbers. Before you can add them you need a precise way to talk about an endless list of numbers and where it is heading. That list is a sequence, and this lesson builds its whole toolkit.

• Find a formula for the general term of a sequence, explicit or recursive
• Say exactly what it means for a sequence to converge, with the epsilon-N definition, and name the ways it can diverge
• Compute limits with the limit laws, by borrowing from a function (L'Hôpital included), and through a continuous function
• Use the squeeze theorem on a sequence with an oscillating factor
• Prove convergence without finding the limit, using the Monotone Convergence Theorem

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, pp. 427-449 — learning objectives 5.1.1 to 5.1.3

This chapter asks what it could possibly mean to add up infinitely many numbers. You will answer that in Section 5.2 by adding the numbers one at a time and watching where the running totals go. Those running totals form an endless list, a sequence, so the first job is to understand sequences on their own.

Most of the machinery is familiar. Limits at infinity, the limit laws, L'Hôpital's rule and the squeeze theorem all come from Calculus I, and this lesson shows how each one carries over when the input is only allowed to be a whole number. The one genuinely new tool is the last objective: a theorem that proves a sequence converges without ever telling you its limit.

That last idea will sound strange at first. By the end of the chapter it will be the idea everything else rests on, so watch for it.

3. Before anything new: limits at infinity

Warm-up

Write both answers down before you reveal them. These two limits are the two standard moves from Calculus I, and both will reappear within the next twenty slides applied to sequences.

For the rational function, dividing the top and the bottom by the highest power of x in the denominator turns every other term into something that visibly goes to zero, leaving three over one. For the logarithm over x, both parts grow without bound, which is the infinity over infinity form, so L'Hôpital's rule applies: differentiate the top and the bottom separately, and one over x on top goes to zero.

If either one felt rusty, that is worth fixing now, because in this lesson the only new thing is the kind of input. The algebra is the same.

4. Sequences and their terms

Section

Part 1

5. An ordered list, given by a formula or a rule

Concept

infinite sequence — An ordered list of numbers, one for each index n. Each number is a term; n is the index variable.

\[ a_1,\; a_2,\; a_3,\; \ldots,\; a_n,\; \ldots \qquad \text{written } \{a_n\}_{n=1}^{\infty} \text{ or } \{a_n\} \]

The list 2, 4, 8, 16, 32 and so on can be given two ways. An explicit formula says what the nth term is outright; a recurrence relation says how to get each term from the one before.

\[ \text{explicit: } a_n = 2^n \qquad \text{recursive: } a_1 = 2,\; a_n = 2a_{n-1} \; (n \ge 2) \]

The braces look like set notation, but order matters here and repeats are allowed. The index may start at 0 or anywhere else.

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 428 — definition of an infinite sequence

A sequence is simply an infinite list with a first entry, a second entry, and so on forever. The subscript is the address of each entry. The notation with braces is borrowed from sets, but a sequence is not a set: the order matters, and the same number may appear many times.

There are two ways to hand someone a sequence. An explicit formula lets you jump straight to any term: the hundredth term of two to the n is two to the hundred, no questions asked. A recurrence relation gives you the starting term and a rule for producing each term from the one before, so to get the hundredth term you would have to walk through the first ninety-nine.

Both descriptions of the powers of two define exactly the same list. Much of the first part of this lesson is about converting a recurrence into an explicit formula, because the explicit formula is what you need to find a limit.

6. A sequence is a function on the whole numbers

Concept

There is exactly one term for each positive integer n, so a sequence is a function whose domain is the positive integers. Its graph is the set of points whose coordinates are n and the nth term.

\[ a: \{1, 2, 3, \ldots\} \to \mathbb{R}, \qquad n \mapsto a_n \]

That graph is dots, never a curve: the sequence has no value at three and a half. Keep this picture; every idea in the lesson is about where the dots go as n grows.

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, pp. 428-429 — Figure 5.2

Here is the idea that connects sequences to everything you already know. Feed in a whole number n and the sequence hands back one number, the nth term. That is exactly what a function does, except that the allowed inputs are only the positive integers.

So a sequence has a graph, and the picture shows the graph of two to the n. Look at what is not there: nothing is drawn between the dots, because there is no third-and-a-half term. The dots are the whole graph.

It is tempting to join the dots with a smooth curve, and later in the lesson you will deliberately draw such a curve to borrow limits from it. But that curve is a different object, a function of a real variable. Keeping the two apart is the key to understanding why some theorems only work in one direction.

7. Arithmetic sequences: a constant difference

Concept

In 3, 7, 11, 15, 19 each term is the previous one plus 4. Unwind the recurrence to find the explicit formula.

\[ a_1 = 3, \quad a_n = a_{n-1} + 4 \quad (n \ge 2) \]

\[ a_2 = 3 + 4, \quad a_3 = 3 + 2\cdot 4, \quad a_4 = 3 + 3\cdot 4 \]

\[ a_n = 3 + 4(n-1) \]

\[ a_n = 4n - 1 \]

Adding 4 once for each step after the first gives n minus 1 fours. Every arithmetic sequence has this shape, a linear formula in n.

\[ \text{arithmetic: } a_n = cn + b \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 429 — arithmetic sequences

An arithmetic sequence is the simplest kind of recurrence: add the same number every time. The move for finding the explicit formula is worth learning because it works for many recurrences: write out a few terms without simplifying and count.

The second term has one four added, the third has two, the fourth has three. So the nth term has one fewer four than its position number, which is where n minus 1 comes from. Simplifying three plus four times n minus 1 gives four n minus 1.

Check it on the list: n equal to 1 gives 3 and n equal to 5 gives 19. In general, any arithmetic sequence has an explicit formula of the form c times n plus b, a straight-line formula, which is why its dots will lie on a line.

8. Geometric sequences: a constant ratio

Concept

In 2, minus two thirds, two ninths, minus two twenty-sevenths, each term is the previous one times minus one third. Unwind again, now counting multiplications.

\[ a_1 = 2, \quad a_n = -\tfrac13 \cdot a_{n-1} \quad (n \ge 2) \]

\[ a_2 = \left(-\tfrac13\right)2, \quad a_3 = \left(-\tfrac13\right)^{2} 2, \quad a_4 = \left(-\tfrac13\right)^{3} 2 \]

\[ a_n = 2\left(-\frac13\right)^{n-1} \]

The nth term has had n minus 1 multiplications by the ratio. Every geometric sequence is a constant times a power of the ratio; 2 to the n is the case with ratio 2.

\[ \text{geometric: } a_n = c\,r^{n} \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, pp. 429-430 — geometric sequences

A geometric sequence multiplies by the same number every time instead of adding it. The same unwinding trick works, but now you count multiplications. The second term has been multiplied once by minus one third, the third twice, the fourth three times.

So the nth term is the first term, 2, times minus one third raised to the power n minus 1. Notice that a negative ratio makes the signs alternate automatically: an even number of negative factors gives a positive term, an odd number gives a negative one.

The book writes the general geometric sequence as c times r to the n, while the example came out as a power n minus 1. These are the same family, since r to the n minus 1 is just r to the n divided by r, a constant you can absorb into c. Use whichever makes the first term come out right.

9. Constant differences against constant ratios

Picture it

On the left, equal steps of 4 put the dots on a straight line. On the right, multiplying by minus one third flips the sign every time and cuts the size to a third: the dots alternate sides of the axis and collapse toward it.

Compare the two panels. On the left, every step up is the same height, marked in yellow, so the dots march along a straight line and grow without bound. This is the visual signature of adding the same number over and over.

On the right, multiplying by minus one third does two things at once. The sign flips, so the dots land alternately above and below the axis, and the size is cut to a third, so each dot is much closer to the axis than the one before. After only a few steps you can barely see the stems.

Hold on to that right-hand picture. A sequence whose terms flip sign but shrink in size is heading to zero, and you will prove exactly that with the squeeze theorem in Part 4.

10. Example 5.1(a): alternating fractions

Worked example

Find an explicit formula for the nth term.

\[ -\frac12,\; \frac23,\; -\frac34,\; \frac45,\; -\frac56,\; \ldots \]

\[ (-1)^n: \quad -1,\; +1,\; -1,\; +1, \ldots \]

\[ \text{numerator} = n \]

\[ \text{denominator} = n + 1 \]

\[ a_n = \frac{(-1)^n\, n}{n+1} \]

\[ a_1 = \frac{(-1)(1)}{2} = -\frac12, \qquad a_4 = \frac{(+1)(4)}{5} = \frac45 \;\checkmark \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 430 — Example 5.1a

When you are asked for a general term, separate the list into independent pieces and handle each piece on its own. Here there are three: the sign, the numerator and the denominator.

The signs go negative, positive, negative, positive, starting negative at n equal to 1. Minus one to the n does exactly that, because minus one to an odd power is minus one. The numerators count up 1, 2, 3, and the denominators count up 2, 3, 4, always one ahead. Putting the three pieces together gives minus one to the n, times n, over n plus 1.

The check deliberately uses one odd and one even index. The most common mistake with alternating sequences is to get the sign pattern one step out of phase, and a check on only even terms would never catch that. Testing n equal to 1 and n equal to 4 checks both the sign and the fraction.

11. Example 5.1(b): a geometric top over an arithmetic bottom

Worked example

Find an explicit formula for the nth term.

\[ \frac34,\; \frac97,\; \frac{27}{10},\; \frac{81}{13},\; \frac{243}{16},\; \ldots \]

\[ \text{numerator} = 3^n \]

\[ \text{denominator} = 4 + 3(n-1) \]

\[ 4 + 3n - 3 = 3n + 1 \]

\[ a_n = \frac{3^n}{3n+1} \]

\[ a_3 = \frac{27}{10}, \qquad a_5 = \frac{243}{16} \;\checkmark \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 430 — Example 5.1b

Again, split the list and attack the parts separately. The numerators 3, 9, 27, 81 are each three times the one before, so they are powers of three. The denominators 4, 7, 10, 13 each go up by three, so they form an arithmetic sequence.

For the denominators use the unwinding pattern from the arithmetic slide: start at 4 and add three, n minus 1 times. That simplifies to three n plus 1. It is worth doing this with the formula rather than by guessing, because guesses like three n plus 4 are easy to write and wrong by one step.

The check tests two terms that were not used to build the formula. The third and fifth terms both match, which is good evidence that the pattern really is the one you found. Of course, any finite list is consistent with infinitely many formulas; the question is asking for the simplest one.

12. Which ingredient makes which pattern?

Matching

Match each list to the formula that produces it. The goal is to build a mental catalogue, so that when you see alternating signs you immediately think of a power of minus one, and when you see 1, 2, 6, 24 you immediately think factorial.

The two that are easy to confuse are the last pair. In 5, 8, 11, 14 the differences are all three, so it is arithmetic and the formula is linear; the first term fixes the constant, giving three n plus 2. In 3, 6, 12, 24 the ratios are all two, so it is geometric; three times two to the n minus 1 gives 3 at n equal to 1.

Always test a candidate formula on the first term. That single substitution catches most off-by-one mistakes in the exponent or the constant.

13. Checkpoint 5.1: fix the sign and the denominators

Fill the middle

\[ \frac15,\; -\frac17,\; \frac19,\; -\frac{1}{11},\; \ldots \]

\[ a_n = \frac{(-1)^{\,?}}{?} \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 430 — Checkpoint 5.1

Fill in both blanks before checking. This checkpoint looks like Example 5.1(a) but it is built to catch the sign phase: this time the first term is positive.

Minus one to the n would make the first term negative, so you need an exponent that is even when n is 1. Either n plus 1 or n minus 1 works, since both are even exactly when n is odd. For the denominators, 5, 7, 9, 11 go up by two starting at five, so they are five plus two times n minus 1, which is two n plus 3.

Check your answer at n equal to 2: minus one cubed over seven is minus one seventh, which matches the list. If you wrote minus one to the n, that check fails immediately.

14. Example 5.2(a): multiply by minus three each time

Worked example

Find an explicit formula for the recursively defined sequence.

\[ a_1 = 2, \qquad a_n = -3a_{n-1} \quad (n \ge 2) \]

\[ a_2 = -3(2), \quad a_3 = (-3)^2(2), \quad a_4 = (-3)^3(2) \]

\[ a_n = 2(-3)^{n-1} \]

\[ a_n = c\,r^{n-1}, \quad c = 2,\; r = -3 \]

\[ \text{rule: } 2,\,-6,\,18,\,-54 \qquad \text{formula: } 2,\,-6,\,18,\,-54 \;\checkmark \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 431 — Example 5.2a

This recurrence multiplies by minus three every step, so you should expect a geometric sequence. The unwinding technique proves it: write each term in terms of the first without multiplying anything out.

The second term is minus three times 2, the third is minus three times that, so minus three squared times 2, and so on. The exponent on minus three is always one less than the position, which gives 2 times minus three to the n minus 1.

The check compares four terms computed from the rule with four computed from the formula. It is cheap, and it is the right check here, because the formula was built from a guessed pattern and the rule is the actual definition. If they agree on several terms, the pattern is right; strictly, a short induction would prove it for all n.

15. Example 5.2(b): adding smaller and smaller halves

Worked example

\[ a_1 = \frac12, \qquad a_n = a_{n-1} + \left(\frac12\right)^{n} \quad (n \ge 2) \]

\[ a_2 = \frac12 + \frac14 = \frac34 \]

\[ a_3 = \frac34 + \frac18 = \frac78, \qquad a_4 = \frac78 + \frac1{16} = \frac{15}{16} \]

\[ a_n = \frac{2^n - 1}{2^n} \]

\[ a_n = 1 - \frac{1}{2^n} \]

\[ a_n - a_{n-1} = \frac{1}{2^{n-1}} - \frac{1}{2^n} = \frac{1}{2^n} \;\checkmark \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 431 — Example 5.2b

Here the recurrence adds a different amount each time: one half to the n. Work out the first few terms as fractions, not decimals, because fractions show the pattern: one half, three quarters, seven eighths, fifteen sixteenths.

Each denominator is a power of two and each numerator is one less than its denominator. That gives two to the n minus 1, over two to the n, and splitting the fraction gives the cleaner form 1 minus one over two to the n. The picture shows what that means: every term closes half of the remaining gap to 1.

The check is a good habit for any formula you extract from a recurrence. Subtract two consecutive terms of your formula and see whether you get the amount the rule adds. One over two to the n minus 1, minus one over two to the n, is exactly one over two to the n, so the formula satisfies the rule. You will meet this sequence again: it is the running total of a geometric series.

16. Checkpoint 5.2: add six each time

Prediction

\[ a_1 = -4, \qquad a_n = a_{n-1} + 6 \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 431 — Checkpoint 5.2

Pick an option before revealing. The rule adds six at every step, so this is an arithmetic sequence, and its explicit formula must be linear in n. That alone rules out the option with a power.

Among the linear options, test the first term. At n equal to 1 the formula must give minus four. Six n minus 10 gives minus four; minus four plus six n gives two, one six too many, because it adds a six even at the first term.

The reliable way to build such formulas is the one from the arithmetic slide: the first term plus n minus 1 copies of the difference. Minus four plus six times n minus 1 simplifies to six n minus 10.

17. What it means to converge

Section

Part 2

18. Four sequences, four behaviours

Picture it

Read each panel as n grows. In (a) the dots climb forever. In (b) they close in on 1. In (c) they flip between two values and never choose. In (d) they flip sides too, but the flips shrink, and the dots close in on 0.

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, pp. 431-432 — Figure 5.3

These four panels are the book's Figure 5.3, and they show every kind of long-run behaviour you need to recognise. Look at each one and ask a single question: as n grows, do the dots close in on one height?

Panel (a) climbs steadily, three units per step, and never stops: no single height. Panel (b) rises and crowds against the dashed line at 1: yes, one height. Panel (c) jumps between minus one and one forever: the dots are always near one of two heights, never one. Panel (d) is the subtle one: it alternates in sign like (c), but the jumps shrink, and the dots crowd in on zero.

The lesson of (c) against (d) is that alternating signs do not decide anything on their own. What decides convergence is whether the distance to a single number shrinks to nothing.

19. Convergent and divergent sequences

Concept

If the terms become arbitrarily close to one finite number L once n is large enough, the sequence converges and L is its limit. Otherwise it diverges.

\[ \lim_{n\to\infty} a_n = L \qquad \text{or} \qquad a_n \to L \]

\[ 1 - \left(\tfrac12\right)^n \to 1, \qquad \frac{(-1)^n}{n} \to 0 \]

\[ \{1 + 3n\} \text{ and } \{(-1)^n\} \text{ diverge} \]

Only the tail matters. Put any finite number of extra terms in front of a sequence and neither its convergence nor its limit changes.

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, pp. 432-433 — informal definition of convergence

The informal definition says a sequence converges to L when its terms get as close as you like to L once n is large enough. The limit L must be a finite real number. Anything else, running off to infinity or wandering without settling, counts as divergence.

Two of the four sequences from the previous slide converge, one to 1 and one to 0, and two diverge. You can write the limit with the limit symbol or with an arrow; both mean the same thing.

The last sentence on the slide has a practical consequence you will use all the time. Convergence is a property of the tail. You may ignore the first thousand terms of any sequence when deciding whether it converges, and you may add or delete finitely many terms at the front without changing the limit. That is why so many theorems below say for n at least N.

20. The formal definition: a band of half-width epsilon

Concept

The phrases arbitrarily close and large enough are made exact by a challenge and a response. The challenge is a tolerance epsilon; the response is an index N past which every term is within epsilon of L.

\[ a_n \to L \iff \text{for every } \varepsilon > 0 \text{ there is an } N \text{ with } |a_n - L| < \varepsilon \text{ for all } n \ge N \]

\[ a_n = 1 + \frac{2(-1)^n}{n}: \quad |a_n - 1| = \frac2n < 0.25 \iff n > 8, \quad N = 9 \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 433 — definition and Figure 5.4

The informal words close as you like and large enough hide a game with two players. Someone challenges you with a tolerance epsilon, as small as they like. You must respond with an index N such that every term from N on lies within epsilon of L. The sequence converges to L exactly when you can always respond, whatever epsilon they pick.

The picture plays one round of the game. The challenge is epsilon equal to 0.25, drawn as the yellow band from 0.75 to 1.25 around L equal to 1. The first eight dots include several that lie outside the band. From the ninth on, every dot is inside, so N equal to 9 is a valid response.

The last line does the round algebraically: the distance to 1 is two over n, which is below a quarter exactly when n is more than 8. Notice that the eighth term lands exactly on the edge of the band, which is not strictly inside, so N has to be 9, not 8.

21. Reading the epsilon-N definition

Notation

Step through the four annotations and connect each one to the band picture. The for-every-epsilon part is what makes the definition strong: it is not enough to get close once, or close to a fixed tolerance, you must be able to meet every tolerance.

The there-exists-N part is what makes it forgiving. N can be as large as it needs to be, and it can change when epsilon changes. Slowly converging sequences need huge N for small epsilon, and they still converge.

The order of the words matters. Epsilon is chosen first and N is chosen after, in response to it. If you swapped them, demanding one N that worked for every epsilon at once, only sequences that are eventually constant would qualify.

22. Finding N: one plus one over n, within 0.01

Worked example

Show that the terms of the sequence below get within 0.01 of 1, find the first index N that works, then do it for every epsilon.

\[ a_n = 1 + \frac1n \]

\[ |a_n - 1| = \left|\frac1n\right| = \frac1n \]

\[ \frac1n < 0.01 \]

\[ n > \frac{1}{0.01} = 100 \]

\[ N = 101 \]

\[ \frac1n < \varepsilon \iff n > \frac1\varepsilon, \quad N = \left\lfloor \frac1\varepsilon \right\rfloor + 1 \]

\[ a_{100} - 1 = 0.01 \nless 0.01, \qquad a_{101} - 1 = \tfrac{1}{101} \approx 0.0099 < 0.01 \;\checkmark \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 433 — the formal definition, applied

This is the definition used as a calculation. The candidate limit is 1, so the distance from the nth term to the limit is one over n, and you want that distance to be below the tolerance.

Solving one over n below one hundredth gives n above 100, so the first index that works is 101. Then the same three moves are done with a general epsilon in place of 0.01, which is what the definition actually demands: for any epsilon, N is one more than the whole-number part of one over epsilon.

The check looks at the boundary. At n equal to 100 the distance is exactly one hundredth, and the definition asks for strictly less, so 100 fails and 101 succeeds. Checking the index on each side of the boundary is the habit that catches off-by-one errors whenever an index comes out of an inequality.

23. Three ways to diverge

Concept

A divergent sequence fails to settle on a finite number, and it can fail in different ways.

\[ 1 + 3n \to \infty: \quad \text{diverges to infinity} \]

\[ -5n + 2 \to -\infty \text{ as } n \to \infty: \quad \text{diverges to } -\infty \]

\[ (-1)^n = -1, 1, -1, 1, \ldots: \quad \text{diverges by oscillation} \]

Writing that the limit is infinity is shorthand for HOW it diverges; the limit still does not exist. The book prints the second line with n going to minus infinity, a misprint: n always runs to plus infinity.

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, pp. 433-434 — divergence to plus and minus infinity

Divergence is simply the failure to converge, but it is useful to say how a sequence fails. The first two lines show terms that eventually exceed any bound you name, in the positive or negative direction. For those you write that the limit is infinity or minus infinity, as a description of the divergence.

The third line is a different failure. The terms of minus one to the n are perfectly well behaved, always one or minus one, but they never settle on one value, so they diverge by oscillation.

A word about the book: it prints the second example as going to minus infinity as n goes to minus infinity. That is a misprint. The index of a sequence only ever runs toward plus infinity; the terms are what go to minus infinity.

24. Converge, run off, or oscillate?

Sorting

Sort each sequence by what its terms do in the long run, not by what the first few terms look like. Computing the first three or four terms helps, but then think about large n.

Two of these catch people. The sequence n squared minus 10n is negative for the first nine terms, but the n squared term eventually dominates and the terms run off to plus infinity. And minus one to the n times n swings further and further out on both sides; it is unbounded, but it does not go to plus infinity or to minus infinity, because it keeps changing sides. So it belongs with the oscillating ones.

Meanwhile minus one to the n over n alternates in sign yet converges to zero, because the size of the terms shrinks. Alternating is not a reason to diverge.

25. Trap: a limit of infinity is not a limit

Trap

26. Computing limits

Section

Part 3

27. Theorem 5.1: borrow the limit of a function

Concept

If the formula for the terms makes sense for every real x, you can let x run continuously. If the function has a limit at infinity, the dots on its graph share it.

Theorem 5.1 — If the nth term equals f(n) for every n from 1 on, and f(x) tends to a real number L as x tends to infinity, then the sequence converges and its limit is L.

\[ \lim_{x\to\infty} f(x) = L \;\Longrightarrow\; \lim_{n\to\infty} f(n) = L \]

This is what licenses L'Hôpital's rule for sequences: you cannot differentiate with respect to a whole-number n, but you can differentiate f(x).

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 434 — Theorem 5.1

Many sequences are given by a formula that makes sense for every positive real number, not just the whole numbers. The logarithm of n over n is one of them: you can put in 2.5 or pi and get an answer. When that happens you can study the function instead, with all of Calculus I available.

The picture shows why the theorem is true. The dots are points on the curve, at x equal to 1, 2, 3 and so on. If the curve eventually stays within any tolerance of a height L, then so do the dots, since they are on the curve. The curve here falls toward zero, so the sequence does too.

This is the only legitimate way to use L'Hôpital's rule on a sequence. You cannot differentiate with respect to n, because n jumps in steps of one and there is no derivative. You switch to x, differentiate there, find the limit, and then Theorem 5.1 carries the answer back to the sequence.

28. The bridge runs one way only

Counterexample

Before revealing, try to invent your own example. You need a function whose values at the whole numbers settle down while the function itself keeps misbehaving in between.

Sine of pi x is the classic answer. At every whole number its value is exactly zero, so the sequence is zero, zero, zero, and converges to zero. But the function keeps rising to one and falling to minus one between the whole numbers forever, so it has no limit at infinity. The picture shows the dots all sitting on the axis while the curve swings through them.

So the theorem only runs from the function to the sequence. If you find that a sequence converges, you learn nothing about any function that happens to pass through its dots. And if a function has no limit, the sequence might still converge: you have to check it some other way.

29. Geometric sequences with r at least zero

Concept

Theorem 5.1 settles the powers of a fixed non-negative number r at once, using the exponential function r to the x.

\[ 0 \le r < 1: \quad r^x \to 0 \;\Longrightarrow\; r^n \to 0 \]

\[ r = 1: \quad 1^n = 1 \to 1 \]

\[ r > 1: \quad r^n \to \infty \quad \text{(diverges)} \]

For r bigger than one the function limit is infinite, so Theorem 5.1 does not apply as stated, but the terms plainly grow without bound just as r to the x does. Negative r comes after the squeeze theorem.

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 434 — the geometric sequence

Powers of a fixed number r come up constantly in this chapter, so it pays to know their behaviour cold. For r between zero and one, the exponential function r to the x decays to zero, so by Theorem 5.1 the sequence does too. For r equal to one, every term is one.

For r bigger than one, the function grows without bound, and the theorem as stated needs a finite limit L, so strictly it does not apply. But the conclusion is still clear: the terms grow without bound, just as the function does. The picture shows all three cases for r equal to 0.8, 1 and 1.15.

The case of a negative ratio is postponed on purpose. Minus a half to the n cannot be treated as an exponential function of a real x, because a negative number to a fractional power is not a real number. It needs the squeeze theorem, which is Part 4.

30. Theorem 5.2: the algebraic limit laws

Concept

If two sequences converge, sums, multiples, products and quotients of them converge to the matching combination of limits.

\[ \frac1n \to 0 \;\Longrightarrow\; \frac1{n^k} = \underbrace{\frac1n\cdots\frac1n}_{k} \to 0^k = 0 \]

Every law needs both limits to EXIST first. That hypothesis is what goes wrong in the most common mistake of this section.

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 435 — Theorem 5.2

These are the limit laws from Calculus I, restated for sequences. If you know the limits of two sequences, you know the limit of their sum, difference, product, any constant multiple, and their quotient, as long as you are not dividing by zero.

The last line uses the product law repeatedly to extend one over n tending to zero into one over n to the k tending to zero, for any positive whole number k. That single fact, combined with the laws, handles every rational expression in n, as the next examples show.

The warning at the bottom is the one to remember. Every law starts with if the limits exist. When one of the pieces has no limit, such as cosine n, the laws are silent, and using them anyway is the error behind one of this lesson's traps.

31. Why the sum law holds

Concept

The book proves part (iii) with a split epsilon. Give each sequence half the tolerance.

\[ |a_n - A| < \tfrac{\varepsilon}{2} \text{ for } n \ge N_1, \qquad |b_n - B| < \tfrac{\varepsilon}{2} \text{ for } n \ge N_2 \]

\[ N = \max(N_1, N_2) \]

\[ |(a_n + b_n) - (A + B)| \le |a_n - A| + |b_n - B| \]

\[ |(a_n + b_n) - (A + B)| < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon \quad (n \ge N) \]

The second line waits until both sequences are within their half-tolerance; the third is the triangle inequality. That is the whole proof.

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 435 — proof of Theorem 5.2(iii)

It is worth seeing one of the limit laws proved, because the proof shows how the epsilon-N definition is used in practice. The target is to make the sum within epsilon of A plus B.

The trick is to split the tolerance. Ask each sequence to be within half of epsilon of its own limit. Each can do that from its own index onward, and after the larger of the two indices, both are. The triangle inequality says the distance of a sum is at most the sum of the distances, so the total error is at most half plus half, which is epsilon.

Taking the larger of two indices, and splitting epsilon into pieces, are the two standard moves in these proofs. The squeeze theorem's proof, which comes later, uses the same idea with three sequences.

32. Example 5.3(a): five minus three over n squared

Worked example

\[ \lim_{n\to\infty}\left(5 - \frac{3}{n^2}\right) \]

\[ \frac1n \to 0 \]

\[ \frac{1}{n^2} = \frac1n\cdot\frac1n \to 0\cdot 0 = 0 \]

\[ \lim\left(5 - \frac{3}{n^2}\right) = \lim 5 - 3\lim\frac{1}{n^2} \]

\[ = 5 - 3\cdot 0 = 5 \]

\[ a_1 = 2, \quad a_{10} = 4.97, \quad a_{100} = 4.9997 \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, pp. 435-436 — Example 5.3a

This is the limit laws at their most routine. Every piece of the formula is built from constants and one over n, so you start from the one fact you know, one over n goes to zero, and build up.

The product law squares it: one over n squared goes to zero times zero. Then the difference and constant-multiple laws apply because both pieces now have known limits. The limit is five minus three times zero, which is five.

The numerical check uses n equal to 1, 10 and 100. The first term is 2, far from the limit, but by n equal to 100 the term is 4.9997. The first term being far away does not matter at all, which is the tail principle again.

33. Example 5.3(b): a ratio of quartics

Worked example

\[ \lim_{n\to\infty}\frac{3n^4 - 7n^2 + 5}{6 - 4n^4} \]

\[ \frac{3n^4 - 7n^2 + 5}{6 - 4n^4} = \frac{3 - \frac{7}{n^2} + \frac{5}{n^4}}{\frac{6}{n^4} - 4} \]

\[ 3 - 7\cdot 0 + 5\cdot 0 = 3 \]

\[ 6\cdot 0 - 4 = -4 \ne 0 \]

\[ \lim_{n\to\infty}\frac{3n^4 - 7n^2 + 5}{6 - 4n^4} = \frac{3}{-4} = -\frac34 \]

\[ a_{10} \approx -0.73274, \qquad a_{100} \approx -0.74983 \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 436 — Example 5.3b

For a ratio of polynomials the move is the one from the warm-up: divide the top and the bottom by the highest power of n appearing in the denominator, here n to the fourth. Every term then becomes either a constant or a constant over a power of n.

Now the limit laws apply piece by piece. The numerator tends to three and the denominator to minus four. Before using the quotient law, confirm the denominator's limit is not zero; it is minus four, so the law applies and the limit is minus three quarters.

A shortcut you may know: for equal degrees, the limit is the ratio of the leading coefficients, three over minus four. That shortcut is exactly this calculation done in your head. The numerical check at n equal to 10 and 100 shows the terms settling toward minus 0.75.

34. Example 5.3(c): two to the n over n squared

Worked example

\[ \lim_{n\to\infty}\frac{2^n}{n^2} \]

\[ f(x) = \frac{2^x}{x^2}, \quad \frac{\infty}{\infty} \]

\[ \lim_{x\to\infty}\frac{2^x \ln 2}{2x} \quad \left(\text{still } \tfrac{\infty}{\infty}\right) \]

\[ \lim_{x\to\infty}\frac{2^x(\ln 2)^2}{2} = \infty \]

\[ \frac{2^n}{n^2} \to \infty \;\Longrightarrow\; \text{diverges} \]

\[ a_3 = \tfrac89, \quad a_{10} = 10.24, \quad a_{20} = 2621.44 \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 436 — Example 5.3c

Here the limit laws stall immediately: the top and the bottom both grow without bound, and there is no power of n to divide by that tames two to the n. This is the situation Theorem 5.1 is for. Switch to the real variable x, where the form is infinity over infinity and L'Hôpital's rule applies.

One application leaves two to the x times ln 2 over two x, still infinity over infinity, so apply it again. Now the bottom is the constant 2 while the top still grows exponentially, so the limit is infinity. Theorem 5.1 carries that back: the sequence grows without bound, so it diverges.

The picture adds something the algebra hides. For the first three terms the squared denominator actually wins, and the terms fall to eight ninths at n equal to 3. From then on the doubling on top takes over for good. An exponential always beats a polynomial in the end, but not necessarily at the start.

35. Example 5.3(d): the compound-interest limit

Worked example

\[ \lim_{n\to\infty}\left(1 + \frac4n\right)^{n} \]

\[ y = \lim_{x\to\infty}\left(1 + \frac4x\right)^{x} \]

\[ \ln y = \lim_{x\to\infty} x\ln\left(1 + \frac4x\right) \]

\[ \ln y = \lim_{x\to\infty}\frac{\ln(1 + 4/x)}{1/x} \]

\[ \ln y = \lim_{x\to\infty}\frac{\dfrac{-4/x^2}{1 + 4/x}}{-1/x^2} = \lim_{x\to\infty}\frac{4}{1 + 4/x} \]

\[ \ln y = 4 \;\Longrightarrow\; y = e^4 \]

\[ a_{10} \approx 28.93, \quad a_{100} \approx 50.50, \quad a_{10000} \approx 54.55 \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, pp. 436-437 — Example 5.3d

This limit has the form one to the infinity, which is indeterminate: the base is heading to one, which pushes the value toward one, while the exponent grows, which pushes it up. The logarithm turns the exponent into a factor so the two effects can be compared.

Take the log, move it inside the limit because the logarithm is continuous, and pull the exponent down. That produces infinity times zero, which you rewrite as a quotient to get zero over zero. L'Hôpital's rule then gives four over one plus four over x, which tends to four. So the log of the limit is four, and the limit is e to the fourth.

The picture shows how slowly this sequence converges. After ten terms it is only about 29, after a hundred about 50.5, and it takes ten thousand terms to get within a twentieth of 54.598. This is the compound-interest limit: four units of growth compounded n times per period, with n growing, gives continuous compounding.

36. Slide a and watch the limit

Tweak it

This slide deliberately draws a smooth curve, because it shows the function whose values at the whole numbers are the terms of the sequence. By Theorem 5.1, wherever the curve levels off, the sequence levels off too.

Start at a equal to 1 and read off the level: about 2.718, the number e. Move a to 2 and the curve levels off near 7.39, which is e squared. At 3 it approaches about 20.1, e cubed. The pattern is that the limit is e to the a, and the logarithm and L'Hôpital argument from the previous example proves it for every a at once.

Notice also how long the curve takes to flatten when a is large. The bigger a is, the slower the approach, which matches the slow convergence you saw for a equal to 4.

37. Checkpoint 5.3: a polynomial against an exponential

Worked example

\[ \lim_{n\to\infty}\frac{5n^2 + 1}{e^n} \]

\[ f(x) = \frac{5x^2 + 1}{e^x}, \quad \frac{\infty}{\infty} \]

\[ \lim_{x\to\infty}\frac{10x}{e^x} \quad \left(\text{still } \tfrac{\infty}{\infty}\right) \]

\[ \lim_{x\to\infty}\frac{10}{e^x} = 0 \]

\[ \frac{5n^2 + 1}{e^n} \to 0 \]

\[ a_2 \approx 2.842, \quad a_{10} \approx 0.02275, \quad a_{15} \approx 0.00034 \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 437 — Checkpoint 5.3

This is the same technique as Example 5.3(c), with the roles reversed: now the exponential is in the denominator. The form is infinity over infinity, so switch to x and apply L'Hôpital's rule as many times as it takes to reduce the polynomial to a constant.

Two applications do it: the top becomes ten while the bottom is still e to the x, which grows without bound, so the quotient goes to zero. Theorem 5.1 then says the sequence goes to zero too.

The picture confirms it and adds a detail: the terms rise from about 2.21 to about 2.84 at n equal to 2 before collapsing. As before, the early terms say nothing about the limit. The rule to remember is that e to the n beats any polynomial in n, which you can now prove in one line whenever you need it.

38. Theorem 5.3: limits pass through continuous functions

Concept

Theorem 5.3 — If a sequence tends to L and f is continuous at L, then f applied to the terms tends to f(L).

\[ a_n \to L,\; f \text{ continuous at } L \;\Longrightarrow\; f(a_n) \to f(L) \]

\[ 5 - \frac3{n^2} \to 5 \;\Longrightarrow\; \sqrt{5 - \frac3{n^2}} \to \sqrt5 \]

The proof chains two tolerances: continuity turns an output tolerance epsilon into an input tolerance delta, and convergence of the inputs supplies the N for that delta.

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, pp. 437-438 — Theorem 5.3 and Figure 5.5

This theorem lets you take a limit through a function. If the inputs converge to L and the function is continuous at L, then the outputs converge to the function's value at L. You used this in Calculus I for functions of x; here the inputs are the terms of a sequence.

The picture follows the book's example. The inputs are five minus three over n squared, marked on the x-axis, closing in on 5. The dashed paths carry each input up to the square-root curve and across to the y-axis, where the outputs close in on the square root of 5. Continuity is what guarantees the outputs cannot jump away as the inputs settle.

In practice you use it constantly and almost without noticing: to take a limit of the square root of something, the cosine of something, or e to the something, find the limit of the something and apply the function. You did exactly this in Example 5.3(d) when you moved the logarithm inside the limit.

39. Example 5.4: cosine of three over n squared

Worked example

\[ \lim_{n\to\infty}\cos\left(\frac{3}{n^2}\right) \]

\[ \frac{3}{n^2} = 3\cdot\frac{1}{n^2} \to 3\cdot 0 = 0 \]

\[ \cos x \text{ continuous at } x = 0 \]

\[ \cos\left(\frac{3}{n^2}\right) \to \cos 0 = 1 \]

\[ a_1 = \cos 3 \approx -0.98999, \quad a_3 \approx 0.94496, \quad a_{10} \approx 0.99955 \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 438 — Example 5.4

The sequence is a continuous function, cosine, applied to another sequence, three over n squared. That structure tells you the plan: find the inside limit, then pass it through the cosine using Theorem 5.3.

The inside goes to zero by the multiple law. Cosine is continuous everywhere, so in particular at zero, and cosine of zero is one. That is the whole solution.

The check shows why the first term should never be trusted. At n equal to 1 the term is cosine of 3, about minus 0.99, nearly the opposite of the limit. By n equal to 3 it is already about 0.945, and by n equal to 10 it is 0.9996. The early terms tell you almost nothing; the tail tells you everything.

40. Checkpoint 5.4: what is wrapped around what?

Step zero

\[ a_n = \sqrt{\frac{2n+1}{3n+5}} \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 438 — Checkpoint 5.4 (the radical is lost in the extracted text)

Answer the structural question before you compute. Identifying the outside function and the inside sequence is the step that decides which theorem you use, and it is the step people skip.

The outside is the square root and the inside is the fraction two n plus 1 over three n plus 5. The inside is a rational expression, so divide by n and use the limit laws to get two thirds. The square root is continuous at two thirds, so Theorem 5.3 passes the limit through, giving the square root of two thirds, about 0.8165.

One note about the source: the extracted text of this checkpoint lost its square-root sign, as radicals often do. Placed right after Theorem 5.3 it only makes sense with the root. Without it, the answer would simply be two thirds, and the terms at n equal to 100 and 1000, about 0.812 and 0.816, confirm the root version.

41. The squeeze theorem

Section

Part 4

42. Theorem 5.4: trapped between two sequences

Concept

Some terms contain a factor with no limit at all, such as sine n or cosine n, so none of the laws apply. If you can trap the terms between two sequences that share a limit, you still win.

\[ a_n \le b_n \le c_n \text{ for } n \ge N, \qquad a_n \to L, \quad c_n \to L \]

\[ \Longrightarrow\quad b_n \to L \]

Why: once both outer sequences are within epsilon of L, the middle term, stuck between them, is too.

\[ -\varepsilon < a_n - L \le b_n - L \le c_n - L < \varepsilon \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 439 — Theorem 5.4 and Figure 5.6

Some sequences contain a factor that has no limit at all, like sine n or cosine n, which wander between minus one and one forever. The limit laws cannot touch them. The squeeze theorem gets around this by not computing the wobbly sequence at all: it traps it.

In the picture, the green dots wander irregularly, because sine n does. But each one sits between an orange dot above, one plus one over n, and a purple dot below, one minus one over n. Both of those head to 1, and the gap between them closes, so the green dots have nowhere to go but 1.

The last line is the proof in one inequality. Once the lower and upper sequences are both within epsilon of L, the middle term, being between them, is within epsilon of L as well. The bounds only need to hold from some index on, as usual.

43. Reading the squeeze theorem

Notation

Go through the notes one at a time. The first thing to notice is that you have to supply the two outer sequences yourself. The theorem does not tell you what they are; it only tells you what happens once you have found them.

Almost always the bounds come from the range of a wobbly factor: sine and cosine are between minus one and one, and minus one to the n is plus or minus one. Multiply or divide those simple inequalities by whatever else is in the formula, keeping track of signs, and you have your bounds.

The third note is the one that makes or breaks an application. The two bounds must go to the same limit. If your bounds go to minus one and one, the sequence is only trapped in an interval of width two, and you have learned nothing about its limit.

44. Example 5.5(a): cosine n over n squared

Worked example

\[ \lim_{n\to\infty}\frac{\cos n}{n^2} \]

\[ \lim_{n\to\infty}\cos n \text{ does not exist} \]

\[ -1 \le \cos n \le 1 \]

\[ -\frac{1}{n^2} \le \frac{\cos n}{n^2} \le \frac{1}{n^2} \]

\[ -\frac{1}{n^2} \to 0, \qquad \frac{1}{n^2} \to 0 \]

\[ \frac{\cos n}{n^2} \to 0 \]

\[ a_1 \approx 0.5403, \quad a_3 \approx -0.1100, \quad a_{10} \approx -0.0084 \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, pp. 439-440 — Example 5.5a

Start by recognising why the direct approach fails: the cosine of n has no limit, so the quotient law has no numerator limit to work with. That is the signal to squeeze.

The bounds come from the cosine's range, minus one to one. Divide everything by n squared, which is positive, so the inequality signs keep their direction. Now the outer sequences are plus and minus one over n squared, and both go to zero, so the middle one does too.

In the picture, the orange and purple dots are the bounds, collapsing toward the axis, and the green dots on stems are the actual terms, bouncing around between them. You never had to know where each green dot lands, only that it is fenced in.

45. Example 5.5(b): minus one half to the n

Worked example

\[ \lim_{n\to\infty}\left(-\frac12\right)^{n} \]

\[ \left(-\frac12\right)^{n} = (-1)^n\left(\frac12\right)^{n} \]

\[ -\frac{1}{2^n} \le \left(-\frac12\right)^{n} \le \frac{1}{2^n} \]

\[ \pm\frac{1}{2^n} \to 0 \]

\[ \left(-\frac12\right)^{n} \to 0 \]

\[ -0.5,\; 0.25,\; -0.125,\; 0.0625,\; -0.03125,\; \ldots \to 0 \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 440 — Example 5.5b

This sequence could not be handled with Theorem 5.1, because minus one half to a fractional power is not a real number. But it separates into a sign and a size: minus one to the n, times one half to the n.

The sign is plus or minus one, so the whole term lies between minus one half to the n and plus one half to the n. Both bounds are geometric with ratio one half, so both go to zero, and the squeeze theorem does the rest.

The picture shows something a little unusual: every term lies exactly on one of its bounds, on the upper one when n is even and the lower one when n is odd. That is fine; the theorem allows equality. The same argument works for any ratio between minus one and zero, which completes the story of geometric sequences.

46. Complete the geometric-sequence table

Comparison

Example 5.5(b) works for any ratio between minus one and zero. Fill in the verdict for every r.

\[ r^n \to 0 \iff |r| < 1; \quad r^n \to 1 \iff r = 1; \quad \text{otherwise } \{r^n\} \text{ diverges} \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 440 — equations 5.1 to 5.4

Fill in each verdict before checking. This table is the complete summary of the powers of a fixed number r, equations 5.1 to 5.4 in the book, and you will use it constantly in the rest of the chapter.

The pattern is that only the size of r matters for convergence to zero: whenever r is strictly between minus one and one, the powers shrink to zero, whether they alternate or not. Exactly one other value, r equal to one, converges, to one.

Every other ratio diverges, but in different ways. Bigger than one grows without bound. Exactly minus one flips between one and minus one. Less than minus one flips and grows at the same time. When you meet geometric series in Section 5.2, this table becomes the condition for a geometric series to have a sum.

47. Checkpoint 5.5: two n minus sine n, over n

Worked example

\[ \lim_{n\to\infty}\frac{2n - \sin n}{n} \]

\[ \frac{2n - \sin n}{n} = 2 - \frac{\sin n}{n} \]

\[ -\frac1n \le \frac{\sin n}{n} \le \frac1n \]

\[ \frac{\sin n}{n} \to 0 \]

\[ 2 - \frac{\sin n}{n} \to 2 - 0 = 2 \]

\[ a_{10} \approx 2.0544, \quad a_{100} \approx 2.0051, \quad a_{1000} \approx 1.9992 \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 440 — Checkpoint 5.5

Before squeezing, simplify. Splitting the fraction gives two minus sine n over n, and the two is harmless. The only difficult part is sine n over n, and that is a classic squeeze.

Sine n is between minus one and one, so dividing by the positive n puts sine n over n between minus one over n and one over n, and both of those go to zero. With that piece settled, the difference law applies and the limit is two minus zero, which is two.

Notice the order: the squeeze was used on one piece first, and only then did the limit laws apply to the whole, because only then did every piece have a limit. The numerical check shows the wobble around 2 shrinking in size like one over n.

48. Trap: the quotient law on a factor with no limit

Trap

49. Find the error: L'Hôpital on sine n over n

Error analysis

Look for the first line that is not justified before you reveal the annotations. The trouble starts at the very first equals sign.

L'Hôpital's rule has a hypothesis: the quotient must have the form zero over zero or infinity over infinity. Sine of x over x has a bounded top and an infinite bottom, which is not one of those forms, so the rule does not apply. And even when the rule does apply, it says only that if the ratio of derivatives has a limit, the original ratio has the same limit. A ratio of derivatives with no limit tells you nothing.

Two invalid steps produced a confident, false conclusion. The correct answer, from the squeeze, is that sine n over n tends to zero. Check the form before using L'Hôpital, every time.

50. Bounded and monotone sequences

Section

Part 5

51. Bounded above, bounded below

Concept

bounded — A sequence is bounded above if some number M is at least every term, bounded below if some M is at most every term, and bounded if both.

\[ a_n \le M \text{ for all } n \quad (\text{above}), \qquad M \le a_n \text{ for all } n \quad (\text{below}) \]

\[ 0 \le \frac1n \le 1: \quad \text{bounded} \]

\[ 2^n \ge 2 \text{ but no } M \text{ exceeds every } 2^n: \quad \text{unbounded} \]

A sequence that is not bounded is unbounded: some terms are arbitrarily large in size, above or below.

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, pp. 440-441 — bounded sequences

Boundedness is about whether all the terms fit between two horizontal lines. A sequence is bounded above if one number sits at or above every term, bounded below if one number sits at or below every term, and bounded if both.

One over n is bounded: every term is between zero and one. Two to the n is bounded below, by 2, since it only increases, but no number sits above all of its terms, so it is unbounded.

Keep in mind that a bound is just some number that works. If 1 is an upper bound, so is 2, so is a million. Proving a sequence is bounded never requires the best bound, only a valid one, and that flexibility is what makes the next theorem easy to use.

52. Theorem 5.5: convergent sequences are bounded

Concept

Take epsilon equal to 1 in the definition. From N on every term is within 1 of L; before N there are only finitely many terms.

\[ n \ge N: \quad |a_n| \le |a_n - L| + |L| < 1 + |L| \]

\[ |a_n| \le M = \max\left(|a_1|, \ldots, |a_{N-1}|,\; 1 + |L|\right) \quad \text{for all } n \]

Theorem 5.5 — If a sequence converges, then it is bounded.

Read it backwards and it is a divergence test: an unbounded sequence cannot converge. It does NOT say that a bounded sequence converges.

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 441 — Theorem 5.5

The proof is a nice use of the definition with a specific tolerance. Choose epsilon equal to one. Convergence gives an index N after which every term is within one of L, so its size is less than one plus the size of L. That handles the infinite tail with a single bound.

What is left is the finite list of terms before N, and any finite list has a largest size. Take the largest of those sizes and one plus the size of L, and you have a bound for every term of the sequence.

The theorem is most useful turned around. If a sequence is unbounded, it cannot converge, which gives you a quick divergence argument. What the theorem does not say, and the next slide makes this point, is that bounded sequences converge.

53. Trap: bounded does not mean convergent

Trap

54. Increasing, decreasing, eventually

Concept

monotone — A sequence is increasing from some index on if each term is at most the next, decreasing if each is at least the next, and monotone if it is one or the other.

\[ \text{increasing: } a_n \le a_{n+1} \qquad \text{decreasing: } a_n \ge a_{n+1} \qquad (n \ge n_0) \]

\[ 2, 0, 3, 0, 4, 0, 1, -\tfrac12, -\tfrac13, -\tfrac14, \ldots \]

The book's example does anything at all for seven terms, then increases from the eighth. That is eventually increasing, and eventually is all that matters.

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 441 — monotone sequences

A sequence is increasing if each term is at least as big as the one before, and decreasing if each is at most as big. Monotone means one or the other. The book counts constant stretches as allowed in both, which is why the definitions use at most and at least rather than strict inequalities.

The picture shows the book's example of eventually increasing. The first seven terms, in orange, jump around with no pattern at all. From the eighth term, minus one half, the green dots rise steadily: minus a half, minus a third, minus a quarter, and so on toward zero.

Because convergence depends only on the tail, eventually monotone is as good as monotone. You may discard any finite number of misbehaving terms at the start. The theorems below all say for n at least some starting index for exactly this reason.

55. Theorem 5.6: the Monotone Convergence Theorem

Concept

Monotone Convergence Theorem — If a sequence is bounded and monotone from some index on, then it converges.

\[ a_n = \frac{n}{n+1}: \quad \frac{n}{n+1} < \frac{n+1}{n+2}, \quad \frac{n}{n+1} < 1 \;\Longrightarrow\; \text{converges} \]

An increasing sequence cannot oscillate, so it either runs off to infinity or converges; the upper bound forbids running off. The proof uses the completeness of the real numbers and is beyond the book.

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, pp. 441-442 — Theorem 5.6 and Figure 5.7

Here is the theorem the whole chapter will lean on. A sequence that is bounded and eventually monotone must converge. You do not need to know, or even be able to find, what it converges to.

The picture shows why it is believable. An increasing sequence cannot oscillate, so its only options are to run off to infinity or to settle. If there is a ceiling it can never pass, running off is impossible, so it must settle. Notice the two dashed lines: the red one at 1.3 is an upper bound, but the terms settle at the yellow line at 1, the lowest possible ceiling.

The proof needs a property of the real numbers called completeness, which says there are no gaps in the number line for a sequence to fall through. That property is beyond this course, so the book takes the theorem as given, and so will you.

56. Reading the Monotone Convergence Theorem

Notation

The first note says which bound actually does the work. An increasing sequence is automatically bounded below by its first term, so all you ever need to prove is the upper bound. For a decreasing sequence it is the other way round.

The third note is the conceptual heart of the theorem. It produces existence without a value. That seems like a weakness, but it is precisely what you need when the limit has no formula, which is the typical situation for infinite series.

When you do want the value, there is often a second step. For recursive sequences you can let n grow in the recurrence and solve for the limit, as the next two examples show. That step is only legitimate after the theorem has told you a limit exists.

57. Example 5.6(a): four to the n over n factorial

Worked example

\[ a_n = \frac{4^n}{n!}: \quad 4,\; 8,\; \tfrac{32}{3},\; \tfrac{32}{3},\; \tfrac{128}{15},\; \ldots \]

\[ a_{n+1} = \frac{4^{n+1}}{(n+1)!} = \frac{4}{n+1}\cdot\frac{4^n}{n!} = \frac{4}{n+1}\,a_n \]

\[ \frac{4}{n+1} \le 1 \iff n \ge 3 \;\Longrightarrow\; a_{n+1} \le a_n \]

\[ 0 < a_n \le \tfrac{32}{3} \]

\[ a_n \to L \]

\[ L = \lim a_{n+1} = \lim\frac{4}{n+1}\cdot\lim a_n = 0\cdot L = 0 \]

\[ a_8 \approx 1.6254, \quad a_{10} \approx 0.2890, \quad a_{12} \approx 0.0350 \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, pp. 442-443 — Example 5.6a

Writing out the first few terms shows the terms rise, level off at thirty-two thirds, and then fall. So the sequence is not decreasing from the start, only eventually, and the theorem allows that.

The key move is to relate each term to the one before: the next term is the current term times four over n plus 1. That multiplier is at most one once n is at least three, so from the third term on each term is no bigger than the last. All terms are positive, so zero is a lower bound, and the theorem gives convergence.

To find the limit, call it L and take limits on both sides of the relation. The shifted sequence has the same limit L, because shifting only drops the first term, and four over n plus 1 goes to zero, so L equals zero times L, which is zero. A factorial beats any exponential, just as an exponential beats any polynomial.

58. Example 5.6(b): a recursive sequence that finds 1

Worked example

\[ a_1 = 2, \qquad a_{n+1} = \frac{a_n}{2} + \frac{1}{2a_n} \quad (n \ge 1) \]

\[ \frac{a_n}{2} + \frac{1}{2a_n} = \frac{a_n^2 + 1}{2a_n} \ge 1 \iff a_n^2 - 2a_n + 1 \ge 0 \]

\[ a_n^2 - 2a_n + 1 = (a_n - 1)^2 \ge 0 \]

\[ a_n^2 + 1 \le 2a_n^2 \;\Longrightarrow\; a_{n+1} = \frac{a_n^2 + 1}{2a_n} \le a_n \]

\[ L = \frac{L}{2} + \frac{1}{2L} \]

\[ 2L^2 = L^2 + 1 \;\Longrightarrow\; L^2 = 1 \;\Longrightarrow\; L = 1 \]

\[ 2,\; \tfrac54,\; \tfrac{41}{40},\; \tfrac{3281}{3280} \approx 1.000305,\; \ldots \to 1 \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, pp. 442-444 — Example 5.6b (the book prints n ≥ 2; the rule must start at n = 1)

Computing a few terms, 2, then five quarters, then forty-one fortieths, suggests the sequence is decreasing toward 1. The proof establishes the lower bound first, because the decreasing step uses it.

For the lower bound, write the next term as a single fraction and compare with 1. That comparison reduces to a sub n minus 1, all squared, being at least zero, which is always true. For the decreasing step, the lower bound gives 1 at most a sub n squared, and adding a sub n squared to both sides and dividing by two a sub n shows the next term is at most the current one. Decreasing and bounded below: the theorem says it converges.

Only now may you take limits in the recurrence. The limit satisfies L equals L over two plus one over two L, which gives L squared equals one. Minus one is ruled out because every term is at least 1, so L is 1. The cobweb picture shows the same thing geometrically. A note on the book: it states the rule for n at least 2, which would leave the second term undefined; the rule has to hold from n equal to 1.

59. Checkpoint 5.6: halve it each time

Prediction

\[ a_1 = 1, \qquad a_n = \frac{a_{n-1}}{2} \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 444 — Checkpoint 5.6

Choose an option before revealing. The rule divides each term by two, so starting from 1 every term is positive and smaller than the one before: the sequence is decreasing and bounded below by zero, and the theorem applies.

Setting up the limit equation the same way as Example 5.6(b) gives L equals L over two. Subtract L over two from both sides and L over two is zero, so L is zero. There is only one root, so nothing needs to be discarded this time.

It is worth noticing that this sequence is just one half to the n minus 1, a geometric sequence with ratio one half, so the table from Part 4 already said it goes to zero. Getting the same answer two ways is a good check that the method works.

60. Put the recursive-limit argument in order

Ranking

Arrange the steps before checking. The logic is fixed at one point in particular: you cannot write an equation for L until you know L exists, so the Monotone Convergence Theorem must come before the limit equation.

Why? Because some recurrences have a limit equation with a perfectly good solution even though the sequence diverges. For example, a sub n plus 1 equals two a sub n gives L equals two L, so L is zero, yet starting from 1 the terms are 1, 2, 4, 8 and diverge. Solving for L first would give a confident wrong answer.

The last step matters too. A limit equation is often a quadratic with two roots, and the bounds you already proved decide which one is real.

61. Fibonacci ratios: where do they settle?

Estimation

\[ F_0 = 0,\; F_1 = 1,\; F_n = F_{n-1} + F_{n-2}: \quad 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, \ldots \]

\[ r = 1 + \frac1r \;\Longrightarrow\; r^2 - r - 1 = 0 \;\Longrightarrow\; r = \frac{1 + \sqrt5}{2} \approx 1.618034 \]

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 445 — Student Project: Fibonacci numbers

Make your estimate from the first few ratios before revealing. They are 1, 2, 1.5, 1.667, 1.6, 1.625, and they bounce above and below something between 1.6 and 1.625.

The picture shows them landing alternately on either side of the dashed line at the golden ratio, each time closer. If you assume the ratio has a limit r, you can find it by dividing the Fibonacci rule by the earlier term, which gives r equals one plus one over r. That is a quadratic whose positive root is one plus root five, over two, about 1.618034.

This is the book's student project, and it raises the same caution as the previous slide: the equation only finds r once you know a limit exists. These ratios are not monotone, they alternate, so the Monotone Convergence Theorem does not apply directly. The book's project proves convergence with an explicit formula for the Fibonacci numbers instead.

62. Newton's method is a sequence

Real world

OpenStax Calculus Volume 2, §5.1 Sequences §5.1, p. 448 — Exercise 54

Newton's method from Calculus I is a recurrence: each new estimate is computed from the last. So the question of whether Newton's method works is the question of whether a sequence converges.

For f equal to x squared minus 2 the recurrence simplifies to x over 2 plus one over x. Starting at 1, the terms are 1.5, then 1.416667, then 1.414216, and by the third iteration the first four decimals, 1.4142, have stopped changing. Setting up the limit equation gives L squared equal to 2, and since every term is positive, the limit is the square root of 2.

Compare Example 5.6(b): its rule is the same Newton iteration for the square root of 1. Every square-root button on a calculator runs a sequence like this, and the reason it can stop after a handful of steps is that the correct digits roughly double each time.

63. Putting it together

Section

Part 6

64. Pattern: finding the limit of a sequence

Pattern

1. A formula in n? Try the limit laws (divide by the highest power). If they stall, pass to f(x) and use L'Hôpital; then Theorem 5.1 brings the limit back.
2. A continuous function of a convergent sequence? Take the inside limit and pass it through (Theorem 5.3).
3. A wobbling factor such as sin n, cos n, (−1)ⁿ? Bound it and squeeze (Theorem 5.4).
4. Defined recursively? Prove bounded and monotone, conclude convergence, then solve L = g(L) (Theorem 5.6).
5. Unbounded, or oscillating without shrinking? It diverges (Theorem 5.5).

This is the order of thought for any limit of a sequence. Before you calculate, look at the shape of the formula, because the shape picks the tool.

If the formula is a rational expression, the limit laws are quickest. If it mixes exponentials, logarithms and powers so that the laws stall at infinity over infinity, switch to f of x and use L'Hôpital, remembering that Theorem 5.1 only runs from the function to the sequence. If a continuous function is wrapped around a convergent sequence, pass the limit through it. If there is a wobbly factor with no limit, bound it and squeeze. If the sequence is recursive, prove it is bounded and monotone, and only then solve for the limit.

And if the terms are unbounded, or oscillate without shrinking, stop: the sequence diverges.

65. Check: which sequence converges?

Check

All four options involve minus one to the n, so the alternating sign cannot be what decides this. Look at the size of the terms in each case.

In the correct option the size is one over root n, which shrinks to zero, so the squeeze theorem pins the terms to zero. In the other three the size either stays fixed or grows, and a sequence that keeps jumping by a fixed amount or more cannot settle on one value.

66. Check: the compound-interest limit

Check

The tempting answer is 1, from the reasoning that the base goes to 1 and one to any power is one. That reasoning ignores the exponent, which is growing at the same time, and one to the infinity is an indeterminate form precisely because those two effects compete.

Run the logarithm argument from Example 5.3(d) with 2 in place of 4 and you get e squared. The slider slide showed the general rule: the constant in the numerator becomes the exponent of e.

67. Check: what the Monotone Convergence Theorem gives

Check

The hypotheses are exactly those of the theorem: increasing, and bounded above by 5. So convergence is guaranteed. The question is what you can say about the limit.

The most common wrong answer is that the limit is 5. But 5 is just one number that happens to sit above every term. The limit is the lowest possible ceiling, which can be far below the bound you proved. The sequence n over n plus 1 satisfies every hypothesis with the bound 5 and converges to 1. All you can say is that the limit is at most 5.

68. Why series will need this theorem

Explain it to yourself

Write your two sentences before revealing. This is the bridge to the next section, and seeing it now will make Section 5.3 feel natural rather than surprising.

A series is summed by forming its partial sums: the first term, the first two terms added, the first three, and so on. If every term is positive, each partial sum is the previous one plus something positive, so the partial sums form an increasing sequence. By the Monotone Convergence Theorem, an increasing sequence converges exactly when it is bounded above.

So for series with positive terms, the whole question of convergence reduces to finding an upper bound for the partial sums. That is what the integral test and the comparison tests in the coming sections do, and it is why they can prove convergence without ever finding the sum.

69. Exit ticket

Exit ticket

\[ a_1 = 1, \qquad a_{n+1} = \sqrt{2 + a_n} \]

This problem uses the whole recursive-sequence procedure in one go. Compute a few terms to see what is happening, prove the bound, prove the monotonicity, then find the limit.

The terms 1, 1.732, 1.932, 1.983 are increasing and approaching 2. For the bound, if a term is below 2 then the next one is the root of something below 4, so it is below 2 too; since the first term is 1, every term is below 2. For monotonicity, the difference of the squares of consecutive terms factors as two minus a sub n, times one plus a sub n, which is positive while the terms are below 2, so the terms increase.

The theorem now guarantees a limit, and the limit equation L equals the root of 2 plus L gives a quadratic with roots 2 and minus 1. Every term is positive, so the limit is 2.

70. Recap

Recap

\[ r^n \to 0 \iff |r| < 1, \qquad \text{convergent} \Rightarrow \text{bounded} \;(\text{not conversely}) \]

Next, Section 5.2 turns every infinite sum into a sequence of partial sums, and all of these tools apply to it.

Stewart, Calculus: Early Transcendentals 8e, §11.1 Sequences §11.1, pp. 694-706 — the same material in Stewart

Sequences are functions on the whole numbers, and their limits are defined by the epsilon-N game: every tolerance must eventually be met by the whole tail. The tail is all that matters; the first thousand terms never affect convergence.

For computing limits you have four tools. The limit laws, when every piece converges. Theorem 5.1, which lets you work with a function of x and use L'Hôpital, but only in the direction from the function to the sequence. Theorem 5.3, which passes limits through continuous functions. And the squeeze theorem, for anything with a bounded wobbly factor.

Then there is the tool with no Calculus I analogue: bounded plus monotone implies convergent, with no value needed. Section 5.2 applies it to the partial sums of a series, and it will be the foundation of every convergence test that follows.