Convergent and Divergent Series — AP® Calculus BC

Part 1 · Hook

Why this matters

Walk halfway to a wall, then half the remaining distance, then half again. Infinitely many steps, yet you never pass the wall: the distances add to a finite total. Other endless sums grow without bound. Deciding which, and naming the total when there is one, is what Unit 10 is about, and it all rests on one definition.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Does the sequence $a_n = \frac{n}{n + 1}$ converge?

Yes, to 1
Yes, to 0
No: its terms keep growing
No: it alternates

Show the answer

$\frac{n}{n + 1}$ $= \frac{1}{1 + 1/n} \to 1$ .

Correct: Yes, to 1:
Yes, to 0:
No: its terms keep growing:
No: it alternates:

2. $\int_1^{\infty} \frac{1}{x^2}\,dx$ means

$\lim_{b \to \infty} \int_1^b \frac{1}{x^2}\,dx$ , which is 1
$\int_1^{1000} \frac{1}{x^2}\,dx$
$\frac{1}{\infty^2} - 1$
nothing: the region is infinite

Show the answer

An improper integral is a limit of definite integrals; here $1 - \frac{1}{b} \to 1$ .

Correct: $\lim_{b \to \infty} \int_1^b \frac{1}{x^2}\,dx$ , which is 1:
$\int_1^{1000} \frac{1}{x^2}\,dx$ :
$\frac{1}{\infty^2} - 1$ :
nothing: the region is infinite:

Part 4 · See it

See it first

For the series with terms a sub n = 1 over n(n + 1), n = 1 to 12: blue bars show each term, starting at 1/2, then 1/6, 1/12 and shrinking toward 0; emerald dots show each partial sum S sub n = n/(n + 1): 1/2, 2/3, 3/4 and so on, rising toward the dashed line at height 1. The sequence of partial sums has limit 1, so the series converges and its sum is 1. — Terms $\frac{1}{n(n + 1)}$ shrink to 0 while the partial sums $\frac{n}{n + 1}$ close in on 1: the series converges, to 1. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Add the first $n$ terms of $\sum a_n$ the partial sum $S_n$ $= a_1 + a_2 + \cdots + a_n$
Do this for every $n$ a new sequence, $S_1, S_2, S_3, \ldots$
If $\lim_{n \to \infty} S_n = S$ , a finite numberthe series converges and its sum is $S$
If the partial sums grow without bound or never settlethe series diverges and has no sum

Part 6 · Key ideas

Key ideas

An infinite series $\sum_{n=1}^{\infty} a_n$ converges to $S$ exactly when its partial sums $S_n$ approach $S$ . Two sequences are in play: the terms $a_n$ and the partial sums $S_n$ . Keep them apart.
From partial sums back to terms: $a_1 = S_1$ and $a_n = S_n - S_{n-1}$ for $n \ge 2$ .
Say \"the series converges\" or \"the series diverges\". The terms of a convergent series approach 0, not the sum.
Constants factor out of a convergent series; adding or removing finitely many terms changes the sum but never whether the series converges.

Part 7 · Misconception

A common mistake

The wrong idea: The terms of $\sum \ln\left(\frac{n + 1}{n}\right)$ approach 0, so the series converges.

What actually happens: Its partial sums are $\ln(n + 1)$ , which grow without bound: the series diverges. Terms that shrink to 0 are not enough.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

Partial sums of a series

The graph shows the partial sums

S_n = \frac{3n}{n + 2}

of a series

\sum_{n=1}^{\infty} a_n

, for

n = 1

to

10

. The dashed line is at height 3.

1. What is the sum of the series $\sum_{n=1}^{\infty} a_n$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

The sum of a series is the limit of its partial sums: $\lim_{n \to \infty} \frac{3n}{n + 2}$ $= 3$ .

Answer: $3$

2. Find $a_3$ , the third term of the series.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$a_3$ $= S_3 - S_2$ $= \frac{9}{5} - \frac{3}{2}$ $= \frac{3}{10}$ : the third partial sum minus the second.

Answer: $\frac{3}{10}$ ≈ 0.300

3. Which statement about this series is correct?

The terms converge to 3, so the series converges
The series diverges, since $S_n$ keeps growing as $n$ grows
The series converges to 3, and its terms approach 0
The series converges to 3, and its terms approach 3

Show the answer

A series converges when its partial sums have a limit; here $S_n \to 3$ , so the series converges, to 3. Say it about the series: \"the series converges\", not \"the terms converge to 3\".

The terms converge to 3, so the series converges: The terms approach 0, not 3. It is the partial sums that approach 3.
The series diverges, since $S_n$ keeps growing as $n$ grows: Partial sums can rise forever and still approach a limit; here they level off at 3.
Correct: The series converges to 3, and its terms approach 0: Right: the partial sums approach 3, and $a_n$ $= \frac{6}{(n + 1)(n + 2)} \to 0$ .
The series converges to 3, and its terms approach 3: If every term were near 3, the partial sums would grow by about 3 each step and never level off.

Table

A series whose partial sums jump

Terms and partial sums of $\sum_{n=1}^{\infty} (-1)^{n+1}$
$n$	$1$	$2$	$3$	$4$	$5$
$a_n$	$1$	$-1$	$1$	$-1$	$1$
$S_n$	$1$	$0$	$1$	$0$	$1$

4. Does the series $\sum_{n=1}^{\infty} (-1)^{n+1}$ converge?

Yes, to 0, since the even partial sums are 0
Yes, to 1, since the first term is 1
Yes, since each term is 1 or $-1$ and so stays bounded
No: the partial sums jump between 1 and 0

Show the answer

$S_n$ is 1 for odd $n$ and 0 for even $n$ . The sequence of partial sums has no limit, so the series diverges.

Yes, to 0, since the even partial sums are 0: The odd partial sums are 1; the whole sequence of partial sums has no limit.
Yes, to 1, since the first term is 1: The sum of a series is the limit of all its partial sums, not its first term.
Yes, since each term is 1 or $-1$ and so stays bounded: Bounded partial sums are not enough; they must approach a single number.
Correct: No: the partial sums jump between 1 and 0: Right: a series converges when its partial sums approach one number, and these do not.

5. The partial sums of $\sum_{n=1}^{\infty} \ln\left(\frac{n + 1}{n}\right)$ are $S_n = \ln(n + 1)$ . What does this show?

The series converges, since its terms approach 0
The series converges to $\ln 2$ , its first term
The series diverges, though its terms approach 0
The series converges to 0

Show the answer

$S_n$ $= \ln 2 - \ln 1$ $+ \ln 3 - \ln 2$ $+ \cdots$ $= \ln(n + 1) \to \infty$ , so the series diverges, even though each term $\ln\left(1 + \frac{1}{n}\right)$ approaches 0.

The series converges, since its terms approach 0: Terms approaching 0 do not make a series converge: these partial sums grow forever.
The series converges to $\ln 2$ , its first term: $\ln 2$ is $a_1 = S_1$ . The sum would be $\lim S_n$ , which does not exist.
Correct: The series diverges, though its terms approach 0: Right: $\ln(n + 1)$ grows without bound, while $\ln\left(1 + \frac{1}{n}\right) \to 0$ .
The series converges to 0: The terms approach 0; the partial sums grow without bound.

6. For a series $\sum_{n=1}^{\infty} a_n$ , what is the partial sum $S_n$ ?

The $n$ th term, $a_n$
The sum of the first $n$ terms, $a_1 + a_2 + \cdots + a_n$
The sum of the terms after the $n$ th one
The limit of $a_n$ as $n \to \infty$

Show the answer

$S_n$ $= \sum_{k=1}^{n} a_k$ . The series converges to $S$ exactly when $\lim_{n \to \infty} S_n = S$ .

The $n$ th term, $a_n$ : $a_n$ is one term; $S_n$ adds the first $n$ of them.
Correct: The sum of the first $n$ terms, $a_1 + a_2 + \cdots + a_n$ : Right: the series converges when these partial sums have a limit.
The sum of the terms after the $n$ th one: That is what is left over, $S - S_n$ , when the series converges.
The limit of $a_n$ as $n \to \infty$ : That limit describes the terms, not the sum of the first $n$ .

Part 9 · Summary

Summary

A series

\sum a_n

converges to

S

when its partial sums

S_n = a_1 + \cdots + a_n

approach

S

; otherwise it diverges. Find terms from partial sums by subtracting, keep the terms and the partial sums apart, and say what converges: the series.

Part 10 · Up next

What comes next

Part 11 · Connections