Telescoping Series: How to Find the Sum

A telescoping series is a series whose terms can be written as differences, $a_n = b_n - b_{n+1}$ , so that most of each partial sum cancels. The nth partial sum collapses to $S_n = b_1 - b_{n+1}$ , and the sum of the series is $b_1 - \lim_{n \to \infty} b_{n+1}$ , if that limit exists.

Before this lesson: Infinite Series and Partial Sums, Divergence Test and Partial Fraction Decomposition

How a telescoping series collapses

An old spyglass telescope slides shut, leaving only its two ends. A telescoping series does the same thing to its partial sums: the middle pieces cancel, and only the first and last pieces are left.

Look at $\tfrac{1}{1 \cdot 2} + \tfrac{1}{2 \cdot 3} + \tfrac{1}{3 \cdot 4} + \cdots$ . Each term splits into a difference, such as $\tfrac{1}{2 \cdot 3} = \tfrac{1}{2} - \tfrac{1}{3}$ . Written that way, the first four terms are below.

S_4 = \left(1 - \tfrac{1}{2}\right) + \left(\tfrac{1}{2} - \tfrac{1}{3}\right) + \left(\tfrac{1}{3} - \tfrac{1}{4}\right) + \left(\tfrac{1}{4} - \tfrac{1}{5}\right) = 1 - \tfrac{1}{5}

Every −½ cancels the next +½, every −⅓ cancels the next +⅓, and so on. Only 1 and −1/5 survive.

The telescoping series rule

If each term is a difference of consecutive values of some sequence $b_n$ , the partial sums and the sum follow directly.

S_n = \sum_{k=1}^{n} \left(b_k - b_{k+1}\right) = b_1 - b_{n+1}

\sum_{n=1}^{\infty} \left(b_n - b_{n+1}\right) = b_1 - \lim_{n \to \infty} b_n \quad \text{(when the limit is finite)}

If $b_n$ has no finite limit, the series diverges. When the terms are $b_n - b_{n+2}$ , two values survive at the front instead of one: the sum is $b_1 + b_2 - 2\lim b_n$ .

How to sum a telescoping series

If the terms are a fraction, split them with partial fraction decomposition.
Write out the first few terms and the last term or two of $S_n$ .
Cancel the matching pieces and write $S_n$ with the few pieces that are left.
Take the limit of $S_n$ as n → ∞. If it is finite, that is the sum; otherwise the series diverges.

Telescoping series examples

Example 1: The classic telescoping series

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

\dfrac{1}{n(n + 1)} = \dfrac{A}{n} + \dfrac{B}{n + 1}

Set up partial fractions.

1 = A(n + 1) + Bn

Multiply both sides by n(n + 1).

A = 1

Let n = 0: 1 = A(1).

B = -1

Let n = −1: 1 = B(−1).

\dfrac{1}{n(n + 1)} = \dfrac{1}{n} - \dfrac{1}{n + 1}

Write the term as a difference.

S_n = \left(1 - \tfrac{1}{2}\right) + \left(\tfrac{1}{2} - \tfrac{1}{3}\right) + \cdots + \left(\tfrac{1}{n} - \tfrac{1}{n + 1}\right)

Write out the partial sum.

S_n = 1 - \dfrac{1}{n + 1}

Cancel the matching pieces.

\lim_{n \to \infty} \left(1 - \dfrac{1}{n + 1}\right) = 1 - 0 = 1

Take the limit; 1/(n + 1) goes to 0.

Answer: the series converges to 1. The partial sums are 1/2, 2/3, 3/4, 4/5, …, which climb toward 1.

Bar chart of the first 10 partial sums of the series 1/(n(n + 1)). The bars have heights 1/2, 2/3, 3/4 and so on up to 10/11, rising toward a dashed line at height 1 but never reaching it. — The partial sums Sₙ = 1 − 1/(n + 1) climb toward 1, so the series converges to 1.

Example 2: Partial fractions with 2n − 1 and 2n + 1

Find the sum of $\sum_{n=1}^{\infty} \dfrac{6}{(2n - 1)(2n + 1)}$ .

\dfrac{6}{(2n - 1)(2n + 1)} = \dfrac{A}{2n - 1} + \dfrac{B}{2n + 1}

Set up partial fractions.

6 = A(2n + 1) + B(2n - 1)

Multiply both sides by (2n − 1)(2n + 1).

6 = 2A \quad \Rightarrow \quad A = 3

Let n = 1/2, which makes 2n − 1 = 0.

6 = -2B \quad \Rightarrow \quad B = -3

Let n = −1/2, which makes 2n + 1 = 0.

\dfrac{6}{(2n - 1)(2n + 1)} = \dfrac{3}{2n - 1} - \dfrac{3}{2n + 1}

Write the term as a difference.

S_n = \left(3 - 1\right) + \left(1 - \tfrac{3}{5}\right) + \left(\tfrac{3}{5} - \tfrac{3}{7}\right) + \cdots + \left(\tfrac{3}{2n - 1} - \tfrac{3}{2n + 1}\right)

Write out the partial sum (n = 1 gives 3 − 1).

S_n = 3 - \dfrac{3}{2n + 1}

Cancel the matching pieces.

\lim_{n \to \infty} \left(3 - \dfrac{3}{2n + 1}\right) = 3

3/(2n + 1) goes to 0.

Answer: the series converges to 3.

Example 3: Terms that cancel two places apart

Find the sum of $\sum_{n=1}^{\infty} \dfrac{4}{n(n + 2)}$ .

\dfrac{4}{n(n + 2)} = \dfrac{A}{n} + \dfrac{B}{n + 2}

Set up partial fractions.

4 = A(n + 2) + Bn

Multiply both sides by n(n + 2).

A = 2,\ B = -2

Let n = 0, then n = −2.

\dfrac{4}{n(n + 2)} = \dfrac{2}{n} - \dfrac{2}{n + 2}

Write the term as a difference.

S_n = \left(2 - \tfrac{2}{3}\right) + \left(\tfrac{2}{2} - \tfrac{2}{4}\right) + \left(\tfrac{2}{3} - \tfrac{2}{5}\right) + \cdots + \left(\tfrac{2}{n} - \tfrac{2}{n + 2}\right)

Write out the partial sum. Each negative piece cancels a positive piece two terms later.

S_n = 2 + 1 - \dfrac{2}{n + 1} - \dfrac{2}{n + 2}

The first two positive pieces and the last two negative pieces survive.

\lim_{n \to \infty} S_n = 2 + 1 - 0 - 0 = 3

Take the limit.

Answer: the series converges to 3. Check: $S_1 = \tfrac{4}{3}$ , and the formula gives $3 - 1 - \tfrac{2}{3} = \tfrac{4}{3}$ .

Example 4: A telescoping series that diverges

Does $\sum_{n=1}^{\infty} \ln\left(\dfrac{n}{n + 1}\right)$ converge?

\ln\left(\dfrac{n}{n + 1}\right) = \ln n - \ln(n + 1)

Use the quotient property of logarithms (see properties of logarithms).

S_n = (\ln 1 - \ln 2) + (\ln 2 - \ln 3) + \cdots + (\ln n - \ln(n + 1))

Write out the partial sum.

S_n = \ln 1 - \ln(n + 1)

Cancel the matching pieces.

S_n = -\ln(n + 1)

ln 1 = 0.

\lim_{n \to \infty} \left(-\ln(n + 1)\right) = -\infty

ln(n + 1) grows without bound.

Answer: the series diverges. Notice that its terms do go to 0, since ln(n/(n + 1)) → ln 1 = 0, so the divergence test could not decide this one.

Example 5: Water added to a tank (word problem)

During hour n, a pipe adds $\dfrac{30}{(n + 1)(n + 2)}$ liters of water to a tank. How much water is added in the first 8 hours? How much would be added if the pipe ran forever?

\dfrac{30}{(n + 1)(n + 2)} = \dfrac{30}{n + 1} - \dfrac{30}{n + 2}

Use partial fractions: 30(n + 2) − 30(n + 1) = 30.

S_n = \left(\tfrac{30}{2} - \tfrac{30}{3}\right) + \left(\tfrac{30}{3} - \tfrac{30}{4}\right) + \cdots + \left(\tfrac{30}{n + 1} - \tfrac{30}{n + 2}\right)

Write out the partial sum.

S_n = 15 - \dfrac{30}{n + 2}

Cancel; 30/2 = 15.

S_8 = 15 - \dfrac{30}{10} = 12

Substitute n = 8.

\lim_{n \to \infty} \left(15 - \dfrac{30}{n + 2}\right) = 15

30/(n + 2) goes to 0.

Answer: 12 liters are added in the first 8 hours, and the total would approach 15 liters if the pipe ran forever.

Common mistakes

Mistake	Example of the error	Fix
Splitting into two divergent series	$\sum \left(\frac{1}{n} - \frac{1}{n+1}\right) = \sum \frac{1}{n} - \sum \frac{1}{n+1} = \infty - \infty$	Both series diverge, so you cannot split. Work with the partial sums instead.
Keeping only one survivor when terms cancel two apart	Writing $S_n = 2 - \frac{2}{n+2}$ for $\sum \left(\frac{2}{n} - \frac{2}{n+2}\right)$	Write out enough terms to see that two pieces survive at each end.
Assuming every telescoping series converges	“The terms cancel, so $\sum \ln\frac{n}{n+1}$ converges”	Find $\lim S_n$ . Here $S_n = -\ln(n + 1) \to -\infty$ .
Sign errors in partial fractions	$\frac{6}{(2n-1)(2n+1)} = \frac{3}{2n+1} - \frac{3}{2n-1}$	Check one value: at n = 1 the term is 2, and 3/1 − 3/3 = 2.
Ignoring the starting index	Using $b_1$ for a series that starts at n = 2	The first surviving piece is the first term actually in the sum, here $b_2$ .
Forgetting the limit of the leftover piece	Writing the sum as $b_1$ without checking $b_n \to 0$	The sum is $b_1 - \lim b_n$ , which equals $b_1$ only when $b_n \to 0$ .

Related methods

Spotting a telescoping series

Look for a fraction with a factored denominator whose factors differ by a constant, like n(n + 1) or (2n − 1)(2n + 1). Also look for differences of the same function at n and n + 1, like $\sqrt{n + 1} - \sqrt{n}$ , $\ln n - \ln(n + 1)$ or $\arctan(n + 1) - \arctan n$ .

When a series does not telescope

Most series have no simple formula for $S_n$ . For those, use a convergence test instead, starting with the integral test in the next lesson.

Practice problems

Find the sum of $\sum_{n=1}^{\infty} \left(\dfrac{1}{n + 2} - \dfrac{1}{n + 3}\right)$ .
Find the sum of $\sum_{n=1}^{\infty} \dfrac{1}{(n + 1)(n + 2)}$ .
Find the sum of $\sum_{n=2}^{\infty} \dfrac{2}{n^2 - 1}$ .
Find the sum of $\sum_{n=1}^{\infty} \dfrac{5}{(5n - 3)(5n + 2)}$ .
Does $\sum_{n=1}^{\infty} \left(\sqrt{n + 1} - \sqrt{n}\right)$ converge?
Find the sum of $\sum_{n=1}^{\infty} \left(\cos\tfrac{1}{n} - \cos\tfrac{1}{n + 1}\right)$ .
Find the sum of $\sum_{n=1}^{\infty} \left(e^{1/n} - e^{1/(n+1)}\right)$ .
Find the sum of $\sum_{n=1}^{\infty} \dfrac{3}{n^2 + 3n}$ .
Find the sum of $\sum_{n=1}^{\infty} \left(\arctan(n + 1) - \arctan n\right)$ .
A charity receives $\dfrac{600}{n(n + 1)}$ dollars in month n from a gift fund. How much does it receive in the first 11 months, and what total does the gift approach?

Show answers

S_n = \left(\tfrac{1}{3} - \tfrac{1}{4}\right) + \left(\tfrac{1}{4} - \tfrac{1}{5}\right) + \cdots + \left(\tfrac{1}{n + 2} - \tfrac{1}{n + 3}\right)

Write out the partial sum.

S_n = \dfrac{1}{3} - \dfrac{1}{n + 3}

Cancel the matching pieces.

\lim_{n \to \infty} S_n = \dfrac{1}{3}

1/(n + 3) goes to 0.

Answer: 1/3

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

Partial fractions: (n + 2) − (n + 1) = 1.

S_n = \dfrac{1}{2} - \dfrac{1}{n + 2}

Write out the partial sum and cancel.

\lim_{n \to \infty} S_n = \dfrac{1}{2}

1/(n + 2) goes to 0.

Answer: 1/2

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

Factor the difference of squares (see factoring the difference of squares).

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

Partial fractions: (n + 1) − (n − 1) = 2.

S_n = \left(1 - \tfrac{1}{3}\right) + \left(\tfrac{1}{2} - \tfrac{1}{4}\right) + \left(\tfrac{1}{3} - \tfrac{1}{5}\right) + \cdots + \left(\tfrac{1}{n - 1} - \tfrac{1}{n + 1}\right)

Write out the partial sum, starting at n = 2.

S_n = 1 + \dfrac{1}{2} - \dfrac{1}{n} - \dfrac{1}{n + 1}

Pieces cancel two apart, so two survive at each end.

\lim_{n \to \infty} S_n = \dfrac{3}{2}

1/n and 1/(n + 1) go to 0.

Answer: 3/2

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

Partial fractions: (5n + 2) − (5n − 3) = 5.

b_n = \dfrac{1}{5n - 3},\ b_{n+1} = \dfrac{1}{5n + 2}

The second piece is the first piece with n + 1 in place of n.

\sum_{n=1}^{\infty} a_n = b_1 - \lim_{n \to \infty} b_n = \dfrac{1}{2} - 0

Use the telescoping rule; b₁ = 1/(5 − 3).

Answer: 1/2

S_n = \left(\sqrt{2} - \sqrt{1}\right) + \left(\sqrt{3} - \sqrt{2}\right) + \cdots + \left(\sqrt{n + 1} - \sqrt{n}\right)

Write out the partial sum.

S_n = \sqrt{n + 1} - 1

Cancel the matching pieces.

\lim_{n \to \infty} \left(\sqrt{n + 1} - 1\right) = \infty

The square root grows without bound.

Answer: no, it diverges

S_n = \cos 1 - \cos\dfrac{1}{n + 1}

Here bₙ = cos(1/n), so the partial sum collapses to b₁ − bₙ₊₁.

\lim_{n \to \infty} \cos\dfrac{1}{n + 1} = \cos 0 = 1

Cosine is continuous and 1/(n + 1) goes to 0.

\lim_{n \to \infty} S_n = \cos 1 - 1 \approx -0.460

Subtract, using radians.

Answer: cos 1 − 1 ≈ −0.460

S_n = e^{1} - e^{1/(n+1)}

Here bₙ = e^(1/n), so the partial sum collapses to b₁ − bₙ₊₁.

\lim_{n \to \infty} e^{1/(n+1)} = e^0 = 1

The exponent goes to 0.

\lim_{n \to \infty} S_n = e - 1 \approx 1.718

Subtract.

Answer: e − 1 ≈ 1.718

\dfrac{3}{n^2 + 3n} = \dfrac{3}{n(n + 3)}

Factor out the GCF n (see factoring out the GCF).

\dfrac{3}{n(n + 3)} = \dfrac{1}{n} - \dfrac{1}{n + 3}

Partial fractions: (n + 3) − n = 3.

S_n = 1 + \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{n + 1} - \dfrac{1}{n + 2} - \dfrac{1}{n + 3}

Pieces cancel three apart, so three survive at each end.

\lim_{n \to \infty} S_n = 1 + \dfrac{1}{2} + \dfrac{1}{3}

The last three pieces go to 0.

1 + \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{6}{6} + \dfrac{3}{6} + \dfrac{2}{6} = \dfrac{11}{6}

Add with a common denominator (see adding and subtracting fractions).

Answer: 11/6

S_n = (\arctan 2 - \arctan 1) + (\arctan 3 - \arctan 2) + \cdots + (\arctan(n + 1) - \arctan n)

Write out the partial sum.

S_n = \arctan(n + 1) - \arctan 1

Cancel the matching pieces.

\lim_{n \to \infty} S_n = \dfrac{\pi}{2} - \dfrac{\pi}{4}

arctan x levels off at π/2, and arctan 1 = π/4 (see inverse trig functions).

\dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4}

Subtract.

Answer: π/4

10.

\dfrac{600}{n(n + 1)} = \dfrac{600}{n} - \dfrac{600}{n + 1}

Partial fractions: 600(n + 1) − 600n = 600.

S_n = 600 - \dfrac{600}{n + 1}

Write out the partial sum and cancel.

S_{11} = 600 - \dfrac{600}{12} = 600 - 50 = 550

Substitute n = 11.

\lim_{n \to \infty} S_n = 600

600/(n + 1) goes to 0.

Answer: $550 in the first 11 months; the total approaches $600

Telescoping series FAQ

What is a telescoping series?

It is a series whose terms are differences like $b_n - b_{n+1}$ , so the partial sums cancel down to a few pieces. For example, $\sum \frac{1}{n(n+1)} = \sum \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1$ .

How do you know if a series is telescoping?

Try to write each term as a difference of the same expression at n and at a later index. Partial fractions often reveal this when the denominator factors into pieces like n and n + 1.

Does every telescoping series converge?

No. The partial sums collapse to $b_1 - b_{n+1}$ , and the series converges only if $b_n$ has a finite limit. For $\sum (\sqrt{n+1} - \sqrt{n})$ , the partial sums grow without bound.

How do you find the sum of a telescoping series?

Write out $S_n$ , cancel the matching pieces, and take the limit of what is left as n → ∞. If the limit is finite, that is the sum.

Summary

A telescoping series has terms $b_n - b_{n+1}$ , so $S_n = b_1 - b_{n+1}$ .
Its sum is $b_1 - \lim b_n$ when that limit is finite; otherwise the series diverges.
Partial fractions often turn a rational term into a telescoping difference.
When pieces cancel k places apart, k pieces survive at each end.
Never split a telescoping series into two divergent series.

Previous lesson: Divergence Test

Next lesson: Integral Test

Back to the Calculus 2 Index