Introduction to Infinite Series

Can you add up infinitely many numbers and get a finite answer? Sometimes you can: $\tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \dots$ adds up to exactly $1$. Sometimes you can’t: $1 + 1 + 1 + \dots$ just grows. This page sets up the language of sequences and series, defines what “adds up to” means, and gives you the first test for spotting a series that can’t possibly have a sum. Everything in the rest of this unit builds on these ideas.

Key ideas

Sequences

A sequence is an ordered list of numbers $a_1, a_2, a_3, \dots$, usually given by a formula for the $n$th term $a_n$. For example, $a_n = \dfrac{1}{n}$ gives $1, \tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4}, \dots$

A sequence converges to $L$ if its terms get as close to $L$ as you like as $n$ grows:

$$\lim_{n \to \infty} a_n = L .$$

You find these limits exactly like limits at infinity of functions: compare the highest powers, or use L’Hôpital’s rule on the matching function of $x$. If the limit doesn’t exist (or is infinite), the sequence diverges.

Series and partial sums

A series is what you get when you add the terms of a sequence:

$$\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \dots$$

You can’t literally add forever, so you add a bit at a time. The $n$th partial sum is the sum of the first $n$ terms:

$$S_n = a_1 + a_2 + \dots + a_n .$$

The partial sums $S_1, S_2, S_3, \dots$ form a new sequence. So every series comes with two sequences: the terms $a_n$ and the partial sums $S_n$. Keep them apart. Most mix-ups in this unit come from confusing them.

Convergence of a series

The series converges with sum $S$ if its partial sums approach $S$:

$$\sum_{n=1}^{\infty} a_n = S \quad\text{means}\quad \lim_{n \to \infty} S_n = S .$$

If the partial sums have no finite limit, the series diverges.

For $\sum \frac{1}{n(n+1)}$, the terms go to $0$ while the partial sums go to $1$. The sum of the series is $1$.

Telescoping series

Usually there’s no neat formula for $S_n$. One exception is a telescoping series, where each term splits into a difference and almost everything cancels. Writing out the first few terms of $S_n$ shows you what survives. (Telescoping series aren’t named in the AP course description, but they’re a great way to see partial sums in action.)

The nth term test for divergence

If a series converges, its terms must shrink to $0$. (Since $a_n = S_n - S_{n-1}$ and both partial sums approach the same $S$, their difference approaches $0$.) Turned around, this gives a test:

$$\text{If } \lim_{n \to \infty} a_n \ne 0 \text{ (or the limit doesn't exist), then } \sum a_n \text{ diverges.}$$

This test can only prove divergence. If $\displaystyle\lim_{n \to \infty} a_n = 0$, the test tells you nothing. The series might converge, like $\sum \frac{1}{n(n+1)}$, or diverge, like the harmonic series $\sum \frac{1}{n}$ (see p-series).

Two useful facts

• Multiplying every term by a nonzero constant, or adding two convergent series term by term, keeps convergence: $\sum c\,a_n = c\sum a_n$ and $\sum (a_n + b_n) = \sum a_n + \sum b_n$.
• Adding or removing finitely many terms doesn’t change whether a series converges. It does change the sum, so watch the starting index.

Worked examples

Example 1: Limits of sequences

Does each sequence converge? If so, find its limit.

• (a) $a_n = \dfrac{3n^2 - 1}{n^2 + 4n}$
• (b) $a_n = \dfrac{n}{e^n}$
• (c) $a_n = \cos(n\pi)$

Solution.

(a) The highest powers on top and bottom are both $n^2$, so the limit is the ratio of their coefficients: $\displaystyle\lim_{n \to \infty} a_n = \frac{3}{1} = 3$. The sequence converges to $3$.

(b) Use L’Hôpital’s rule on $\dfrac{x}{e^x}$, which has the form $\dfrac{\infty}{\infty}$:

$$\lim_{x \to \infty} \frac{x}{e^x} = \lim_{x \to \infty} \frac{1}{e^x} = 0$$

The sequence converges to $0$.

(c) $\cos(n\pi)$ is $-1, 1, -1, 1, \dots$, so it equals $(-1)^n$. It keeps jumping and never settles, so the sequence diverges.

Example 2: A telescoping series

Find the sum of $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$.

Solution. Split the term (you can check by finding a common denominator):

$$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$

Write out the $n$th partial sum:

$$\begin{aligned} S_n &= \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{n} - \frac{1}{n+1}\right) \\ &= 1 - \frac{1}{n+1} && \text{everything in the middle cancels} \end{aligned}$$

So $\displaystyle\lim_{n \to \infty} S_n = 1 - 0 = 1$. The series converges, and its sum is $1$.

Check: $S_3 = \tfrac{1}{2} + \tfrac{1}{6} + \tfrac{1}{12} = \tfrac{9}{12} = \tfrac{3}{4}$, and the formula gives $1 - \tfrac{1}{4} = \tfrac{3}{4}$. ✓

Example 3: Using the nth term test

What does the $n$th term test say about each series?

• (a) $\displaystyle\sum_{n=1}^{\infty} \frac{n}{2n + 5}$
• (b) $\displaystyle\sum_{n=1}^{\infty} \left(1 + \frac{1}{n}\right)^n$
• (c) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$

Solution.

(a) $\displaystyle\lim_{n \to \infty} \frac{n}{2n + 5} = \frac{1}{2} \ne 0$, so the series diverges by the $n$th term test. You’re adding numbers close to $\tfrac{1}{2}$ forever.

(b) $\displaystyle\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e \ne 0$, so the series diverges by the $n$th term test.

(c) $\displaystyle\lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0$, so the test is inconclusive. You need a different test. (It turns out this series diverges, as you’ll see on the p-series page.)

Example 4: Working from a formula for the partial sums

The partial sums of a series $\sum_{n=1}^{\infty} a_n$ are $S_n = \dfrac{4n}{2n + 1}$. Find $a_1$, $a_5$, and the sum of the series.

Solution. The first partial sum is just the first term: $a_1 = S_1 = \dfrac{4}{3}$.

Each term is the difference of consecutive partial sums, $a_n = S_n - S_{n-1}$:

$$a_5 = S_5 - S_4 = \frac{20}{11} - \frac{16}{9} = \frac{180 - 176}{99} = \frac{4}{99}$$

The sum is the limit of the partial sums:

$$\sum_{n=1}^{\infty} a_n = \lim_{n \to \infty} \frac{4n}{2n + 1} = 2$$

Common mistakes

Mixing up the terms and the partial sums. $\displaystyle\lim_{n \to \infty} a_n$ is about the terms; the sum of the series is $\displaystyle\lim_{n \to \infty} S_n$. In Example 2 the terms go to $0$ but the sum is $1$. Always ask yourself which sequence you’re talking about.

Using the nth term test to prove convergence. “The terms go to $0$, so the series converges” is false: the harmonic series $\sum \frac{1}{n}$ has terms going to $0$ and still diverges. On the AP exam, a limit of $0$ earns you nothing by itself. You must use another test.

Saying a sequence diverges because its series diverges (or the other way round). The sequence $a_n = \frac{n}{2n+5}$ converges (to $\tfrac{1}{2}$), but the series $\sum \frac{n}{2n+5}$ diverges. Both statements are true at once.

Cancelling the wrong terms in a telescoping series. Write out at least the first three terms and the last one or two before cancelling. If the split is $\frac{1}{n} - \frac{1}{n+2}$, two terms survive at the front, not one.

Ignoring the starting index. $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=3}^{\infty} a_n$ either both converge or both diverge, but their sums differ by $a_1 + a_2$.

Practice

1. (Warm-up) Find $S_1$, $S_2$, $S_3$, and $S_4$ for $\displaystyle\sum_{n=1}^{\infty} \frac{3}{10^n}$. What number do the partial sums seem to approach?

Solution

The terms are $0.3, 0.03, 0.003, 0.0003, \dots$, so

$$S_1 = 0.3, \quad S_2 = 0.33, \quad S_3 = 0.333, \quad S_4 = 0.3333$$

The partial sums approach $0.333\ldots = \tfrac{1}{3}$, so the series converges with sum $\tfrac{1}{3}$.

2. (Warm-up) Does the sequence $a_n = \dfrac{5 - 2n}{3n + 1}$ converge? If so, find its limit.

Solution

The highest powers are both $n$, with coefficients $-2$ and $3$:

$$\lim_{n \to \infty} \frac{5 - 2n}{3n + 1} = -\frac{2}{3}$$

The sequence converges to $-\tfrac{2}{3}$.

3. (Warm-up) Show that $\displaystyle\sum_{n=1}^{\infty} \frac{n^2}{n^2 + 1}$ diverges.

Solution

$$\lim_{n \to \infty} \frac{n^2}{n^2 + 1} = 1 \ne 0$$

so the series diverges by the $n$th term test.

4. (Core) Find the sum of $\displaystyle\sum_{n=1}^{\infty} \frac{2}{n(n+2)}$. (Hint: $\dfrac{2}{n(n+2)} = \dfrac{1}{n} - \dfrac{1}{n+2}$.)

Solution

Write out the partial sum:

$$S_n = \left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \dots + \left(\frac{1}{n-1} - \frac{1}{n+1}\right) + \left(\frac{1}{n} - \frac{1}{n+2}\right)$$

Each negative part cancels with a positive part two terms later. What survives is

$$S_n = 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}$$

As $n \to \infty$, the last two pieces go to $0$, so the sum is $1 + \tfrac{1}{2} = \tfrac{3}{2}$.

5. (Core) For each series, say whether the $n$th term test shows divergence or is inconclusive.

• (a) $\displaystyle\sum_{n=1}^{\infty} \frac{\ln n}{n}$
• (b) $\displaystyle\sum_{n=1}^{\infty} \cos\!\left(\frac{1}{n}\right)$
• (c) $\displaystyle\sum_{n=1}^{\infty} n \sin\!\left(\frac{1}{n}\right)$

Solution

(a) By L’Hôpital’s rule, $\displaystyle\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = 0$. The test is inconclusive.

(b) As $n \to \infty$, $\frac{1}{n} \to 0$, so $\cos\!\left(\frac{1}{n}\right) \to \cos 0 = 1 \ne 0$. The series diverges.

(c) Let $t = \frac{1}{n}$, so $t \to 0^+$: $\displaystyle\lim_{t \to 0^+} \frac{\sin t}{t} = 1 \ne 0$. The series diverges.

6. (Core) The partial sums of a series are $S_n = \dfrac{3n - 1}{n + 1}$. Find $a_1$, $a_3$, and the sum of the series.

Solution

$a_1 = S_1 = \dfrac{2}{2} = 1$.

$$a_3 = S_3 - S_2 = \frac{8}{4} - \frac{5}{3} = 2 - \frac{5}{3} = \frac{1}{3}$$

The sum is $\displaystyle\lim_{n \to \infty} \frac{3n - 1}{n + 1} = 3$.

7. (Core) Consider $\displaystyle\sum_{n=1}^{\infty} \big(\ln(n+1) - \ln n\big)$.

• (a) Show that the terms approach $0$.
• (b) Find a formula for $S_n$ and decide whether the series converges.

Solution

(a) $\ln(n+1) - \ln n = \ln\!\left(\dfrac{n+1}{n}\right) = \ln\!\left(1 + \dfrac{1}{n}\right) \to \ln 1 = 0$.

(b) The series telescopes:

$$S_n = (\ln 2 - \ln 1) + (\ln 3 - \ln 2) + \dots + \big(\ln(n+1) - \ln n\big) = \ln(n+1) - \ln 1 = \ln(n+1)$$

Since $\ln(n+1) \to \infty$, the series diverges, even though its terms go to $0$. This is exactly why the $n$th term test can’t prove convergence.

8. (Challenge) Find the sum of $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2 - 1}$.

Solution

Factor and split: $\dfrac{1}{n^2 - 1} = \dfrac{1}{(n-1)(n+1)} = \dfrac{1}{2}\left(\dfrac{1}{n-1} - \dfrac{1}{n+1}\right)$. (Check: $\dfrac{1}{n-1} - \dfrac{1}{n+1} = \dfrac{2}{(n-1)(n+1)}$.)

Starting at $n = 2$:

$$S_n = \frac{1}{2}\left[\left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \dots + \left(\frac{1}{n-1} - \frac{1}{n+1}\right)\right]$$

Two positive pieces survive at the front ($1$ and $\tfrac{1}{2}$) and two negative pieces at the end ($\tfrac{1}{n}$ and $\tfrac{1}{n+1}$), which go to $0$. So the sum is

$$\frac{1}{2}\left(1 + \frac{1}{2}\right) = \frac{3}{4}$$

9. (Challenge) Suppose $\displaystyle\sum_{n=1}^{\infty} a_n = 7$, with $a_1 = 2$ and $a_2 = 1$.

• (a) Find $\displaystyle\lim_{n \to \infty} a_n$ and $\displaystyle\lim_{n \to \infty} S_n$.
• (b) Find $\displaystyle\sum_{n=3}^{\infty} a_n$.
• (c) Does $\displaystyle\sum_{n=1}^{\infty} (a_n + 1)$ converge? Explain.

Solution

(a) The series converges, so its terms must approach $0$: $\displaystyle\lim_{n \to \infty} a_n = 0$. The partial sums approach the sum: $\displaystyle\lim_{n \to \infty} S_n = 7$.

(b) Removing the first two terms removes $2 + 1 = 3$ from the sum: $7 - 3 = 4$.

(c) No. Since $a_n \to 0$, the terms $a_n + 1 \to 1 \ne 0$, so the series diverges by the $n$th term test.