Infinite Series and Partial Sums

An infinite series adds the terms of a sequence forever: a_1 + a_2 + a_3 + \cdots. Its nth partial sum, S_n, adds just the first n terms. If the partial sums approach one finite number S, the series converges and S is its sum. If not, the series diverges.

Before this lesson: Convergence of Sequences, Sigma Notation, Geometric Series and Infinite Geometric Series

What an infinite series means

You cannot add infinitely many numbers one at a time. Instead, keep a running total. After one term you have S_1, after two terms S_2, and so on. These running totals form a new sequence, the sequence of partial sums.

Picture walking toward a wall 1 meter away. First you walk half the distance, then half of what is left, and so on. Your steps are 1/2, 1/4, 1/8, … meters, and your running totals are 0.5, 0.75, 0.875, …. They close in on 1, so \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \cdots = 1.

So the question “what is the sum of this infinite series?” really means “what number do the partial sums approach?” That turns a series question into a sequence question, which you already know how to answer.

Partial sums and the sum of a series

For a series \sum_{n=1}^{\infty} a_n, the nth partial sum and the sum are defined like this.

S_n = \sum_{k=1}^{n} a_k = a_1 + a_2 + \cdots + a_n

\sum_{n=1}^{\infty} a_n = \lim_{n \to \infty} S_n \quad \text{(when this limit exists and is finite)}

If the limit is a finite number, the series converges. If the limit is infinite or does not exist, the series diverges.

Geometric series. A series where each term is r times the one before has a partial sum formula, so its sum is easy to find. With first term a, the series converges only when |r| \lt 1.

S_n = \dfrac{a\left(1 - r^n\right)}{1 - r} \qquad \sum_{n=1}^{\infty} ar^{n-1} = \dfrac{a}{1 - r} \text{ when } |r| \lt 1

Sum and constant multiple rules. If \sum a_n = A and \sum b_n = B both converge, then \sum (a_n + b_n) = A + B and \sum c\,a_n = cA. These are the limit laws applied to partial sums.

Terms from partial sums. Each term is the change in the running total: a_1 = S_1 and a_n = S_n - S_{n-1} for n ≥ 2.

How to decide if an infinite series converges from its partial sums

1. Write the first few partial sums S_1, S_2, S_3 to see the pattern.
2. Find a formula for S_n, if you can (for example, with the geometric series formula).
3. Find \lim_{n \to \infty} S_n using the methods for convergence of sequences.
4. If the limit is a finite number S, the series converges to S. Otherwise it diverges.

Infinite series examples

Example 1: Partial sums of 0.3 + 0.03 + 0.003 + ⋯

Find the first three partial sums and the sum of \sum_{n=1}^{\infty} \dfrac{3}{10^n}.

S_1 = 0.3

S_2 = 0.3 + 0.03 = 0.33

S_3 = 0.33 + 0.003 = 0.333

S_n = \dfrac{0.3\left(1 - 0.1^n\right)}{1 - 0.1}

S_n = \dfrac{0.3\left(1 - 0.1^n\right)}{0.9}

S_n = \dfrac{1}{3}\left(1 - 0.1^n\right)

\lim_{n \to \infty} 0.1^n = 0

\lim_{n \to \infty} S_n = \dfrac{1}{3}(1 - 0) = \dfrac{1}{3}

Answer: S_1 = 0.3, S_2 = 0.33, S_3 = 0.333, and the series converges to 1/3. This is why the repeating decimal 0.333… equals 1/3.

Example 2: Finding the sum and the terms from S_n

The partial sums of a series are S_n = \dfrac{5n}{2n + 1}. Find the sum of the series and a formula for its terms a_n.

\sum_{n=1}^{\infty} a_n = \lim_{n \to \infty} \dfrac{5n}{2n + 1}

\lim_{n \to \infty} \dfrac{5n}{2n + 1} = \lim_{n \to \infty} \dfrac{5}{2 + \frac{1}{n}}

\lim_{n \to \infty} \dfrac{5}{2 + \frac{1}{n}} = \dfrac{5}{2}

a_1 = S_1 = \dfrac{5(1)}{2(1) + 1} = \dfrac{5}{3}

a_n = S_n - S_{n-1}

S_{n-1} = \dfrac{5(n - 1)}{2(n - 1) + 1} = \dfrac{5(n - 1)}{2n - 1}

a_n = \dfrac{5n}{2n + 1} - \dfrac{5(n - 1)}{2n - 1}

a_n = \dfrac{5n(2n - 1) - 5(n - 1)(2n + 1)}{(2n + 1)(2n - 1)}

a_n = \dfrac{(10n^2 - 5n) - (10n^2 - 5n - 5)}{(2n + 1)(2n - 1)}

a_n = \dfrac{5}{(2n + 1)(2n - 1)}

\dfrac{5}{(2(1) + 1)(2(1) - 1)} = \dfrac{5}{3}

Answer: the series converges to 5/2, and a_n = \dfrac{5}{(2n + 1)(2n - 1)} for every n ≥ 1.

Example 3: A geometric series with alternating signs

Find the sum of \sum_{n=1}^{\infty} 6\left(-\tfrac{2}{3}\right)^{n-1} = 6 - 4 + \tfrac{8}{3} - \tfrac{16}{9} + \cdots.

a = 6

r = -\tfrac{2}{3}

|r| = \tfrac{2}{3} \lt 1

S = \dfrac{6}{1 - \left(-\frac{2}{3}\right)}

S = \dfrac{6}{\frac{5}{3}}

S = 6 \cdot \dfrac{3}{5}

S = \dfrac{18}{5}

Answer: the series converges to 18/5 = 3.6. The partial sums are 6, 2, 4.67, 2.89, 4.07, …, landing above and below 3.6 and closing in on it.

Example 4: Two series that diverge

Show that each series diverges: (a) \sum_{n=1}^{\infty} n = 1 + 2 + 3 + \cdots and (b) \sum_{n=1}^{\infty} (-1)^{n+1} = 1 - 1 + 1 - 1 + \cdots.

Part (a)

S_n = 1 + 2 + \cdots + n

S_n = \dfrac{n(n + 1)}{2}

\lim_{n \to \infty} \dfrac{n(n + 1)}{2} = \infty

Part (b)

S_1 = 1,\ S_2 = 0,\ S_3 = 1,\ S_4 = 0

S_n = 1 \text{ for odd } n \text{ and } S_n = 0 \text{ for even } n

\lim_{n \to \infty} S_n \text{ does not exist}

Answer: both series diverge. In (a) the partial sums go to infinity. In (b) they jump between 1 and 0 forever.

Example 5: A bouncing ball (word problem)

A ball is dropped from a height of 12 meters. Each time it hits the floor, it bounces back up to 3/4 of the height it just fell. If it keeps bouncing forever, what total vertical distance does it travel?

12,\ 9,\ 6.75,\ 5.0625,\ \ldots

D = 12 + 2(9 + 6.75 + 5.0625 + \cdots)

9 + 6.75 + 5.0625 + \cdots = \dfrac{9}{1 - \frac{3}{4}}

\dfrac{9}{1 - \frac{3}{4}} = \dfrac{9}{\frac{1}{4}}

\dfrac{9}{\frac{1}{4}} = 36

D = 12 + 2(36)

D = 12 + 72

D = 84

Answer: the ball travels a total of 84 meters, even though it bounces infinitely many times.

Common mistakes

Related methods

Repeating decimals as series

Any repeating decimal is a geometric series. For example, 0.\overline{27} = 0.27 + 0.0027 + \cdots, with a = 0.27 and r = 0.01, so its sum is \dfrac{0.27}{0.99} = \dfrac{3}{11}.

Tests for convergence

Most series do not have a neat formula for S_n. The rest of this unit gives tests that decide convergence without one. The first is the divergence test, which says that if the terms do not go to 0, the series diverges. Some series with nice partial sums, called telescoping series, are coming up too.

Practice problems

1. Find S_1, S_2, S_3 and S_4 for \sum_{n=1}^{\infty} (2n - 1). Does the series converge?
2. Find the sum of \sum_{n=1}^{\infty} 5\left(\tfrac{1}{4}\right)^{n-1}.
3. Find the sum of \sum_{n=0}^{\infty} \left(-\tfrac{1}{2}\right)^n.
4. Does \sum_{n=1}^{\infty} \left(\tfrac{5}{4}\right)^n converge?
5. The partial sums of a series are S_n = 3 - \dfrac{2}{n}. Find the sum and a formula for a_n.
6. Write the repeating decimal 0.\overline{45} as a fraction using a series.
7. Find the sum of \sum_{n=1}^{\infty} \dfrac{2^n + 3^n}{6^n}.
8. Find the sum of \sum_{n=1}^{\infty} \dfrac{3 \cdot 2^n}{5^{n+1}}.
9. For which values of x does \sum_{n=0}^{\infty} \left(\dfrac{x}{3}\right)^n converge? What is its sum?
10. A snail crawls 60 cm in the first hour. Each hour after that, it crawls 2/3 as far as the hour before. How far does it crawl in the first 4 hours, and what total distance does it approach?

S_1 = 1

S_2 = 1 + 3 = 4

S_3 = 4 + 5 = 9

S_4 = 9 + 7 = 16

S_n = n^2

\lim_{n \to \infty} n^2 = \infty

a = 5,\ r = \tfrac{1}{4}

|r| = \tfrac{1}{4} \lt 1

S = \dfrac{5}{1 - \frac{1}{4}}

S = \dfrac{5}{\frac{3}{4}}

S = 5 \cdot \dfrac{4}{3} = \dfrac{20}{3}

a = \left(-\tfrac{1}{2}\right)^0 = 1

r = -\tfrac{1}{2}

S = \dfrac{1}{1 - \left(-\frac{1}{2}\right)}

S = \dfrac{1}{\frac{3}{2}} = \dfrac{2}{3}

r = \tfrac{5}{4}

|r| = \tfrac{5}{4} \ge 1

S_n = \dfrac{\frac{5}{4}\left(1 - \left(\frac{5}{4}\right)^n\right)}{1 - \frac{5}{4}} = 5\left(\left(\tfrac{5}{4}\right)^n - 1\right)

\lim_{n \to \infty} 5\left(\left(\tfrac{5}{4}\right)^n - 1\right) = \infty

\lim_{n \to \infty} \left(3 - \dfrac{2}{n}\right) = 3

a_1 = S_1 = 3 - 2 = 1

a_n = \left(3 - \dfrac{2}{n}\right) - \left(3 - \dfrac{2}{n - 1}\right)

a_n = \dfrac{2}{n - 1} - \dfrac{2}{n}

a_n = \dfrac{2n - 2(n - 1)}{n(n - 1)}

a_n = \dfrac{2}{n(n - 1)}

0.\overline{45} = 0.45 + 0.0045 + 0.000045 + \cdots

a = 0.45,\ r = 0.01

S = \dfrac{0.45}{1 - 0.01} = \dfrac{0.45}{0.99}

S = \dfrac{45}{99}

S = \dfrac{5}{11}

\dfrac{2^n + 3^n}{6^n} = \dfrac{2^n}{6^n} + \dfrac{3^n}{6^n}

\dfrac{2^n}{6^n} + \dfrac{3^n}{6^n} = \left(\tfrac{1}{3}\right)^n + \left(\tfrac{1}{2}\right)^n

\sum_{n=1}^{\infty} \left(\tfrac{1}{3}\right)^n = \dfrac{\frac{1}{3}}{1 - \frac{1}{3}} = \dfrac{1}{2}

\sum_{n=1}^{\infty} \left(\tfrac{1}{2}\right)^n = \dfrac{\frac{1}{2}}{1 - \frac{1}{2}} = 1

\sum_{n=1}^{\infty} \dfrac{2^n + 3^n}{6^n} = \dfrac{1}{2} + 1 = \dfrac{3}{2}

a_1 = \dfrac{3 \cdot 2^1}{5^2} = \dfrac{6}{25}

\dfrac{a_{n+1}}{a_n} = \dfrac{3 \cdot 2^{n+1}/5^{n+2}}{3 \cdot 2^n/5^{n+1}} = \dfrac{2}{5}

S = \dfrac{\frac{6}{25}}{1 - \frac{2}{5}} = \dfrac{\frac{6}{25}}{\frac{3}{5}}

S = \dfrac{6}{25} \cdot \dfrac{5}{3} = \dfrac{2}{5}

r = \dfrac{x}{3}

\left|\dfrac{x}{3}\right| \lt 1

-3 \lt x \lt 3

S = \dfrac{1}{1 - \frac{x}{3}}

S = \dfrac{3}{3 - x}

S_4 = \dfrac{60\left(1 - \left(\frac{2}{3}\right)^4\right)}{1 - \frac{2}{3}}

S_4 = 180\left(1 - \tfrac{16}{81}\right)

S_4 = 180 \cdot \tfrac{65}{81} = \tfrac{1300}{9} \approx 144.4

S = \dfrac{60}{1 - \frac{2}{3}} = 180

Infinite series FAQ

What is a partial sum of a series?

The nth partial sum S_n is the total of the first n terms of the series. The partial sums form a sequence, and the series converges exactly when that sequence converges.

What is the difference between a sequence and a series?

A sequence is a list of numbers, like 1/2, 1/4, 1/8, …. A series is the sum of those numbers, like 1/2 + 1/4 + 1/8 + ⋯, which here equals 1.

Can a series converge if its terms do not go to 0?

No. If the terms do not approach 0, the partial sums keep changing by a noticeable amount, so the series diverges. This is the divergence test. The opposite is not true: terms going to 0 does not guarantee convergence.

How do you find the sum of an infinite series?

Find a formula for the partial sums and take its limit as n → ∞. For a geometric series with |r| \lt 1, the sum is a/(1 − r), where a is the first term.

Summary

• An infinite series adds the terms of a sequence forever; its nth partial sum S_n adds the first n terms.
• The series converges to S when S_n \to S, a finite number. Otherwise it diverges.
• A geometric series converges to a/(1 − r) when |r| \lt 1 and diverges when |r| \ge 1.
• Each term is the change in the running total: a_n = S_n - S_{n-1}.
• Convergent series can be added and multiplied by constants term by term.

Previous lesson: Convergence of Sequences

Next lesson: Divergence Test

Back to the Calculus 2 Index