How do I know if a series will telescope?

A series telescopes if you can rewrite it as a difference of consecutive terms from a sequence, where each term appears positively in one place and negatively in the next. If the series is not already in this form, try using algebraic identities or decomposition to convert it into telescoping form.

What's the difference between a telescoping series and a geometric series?

A geometric series has a constant ratio between consecutive terms and is summed using S = \frac{a(1-r^n)}{1-r}. A telescoping series cancels out middle terms and simplifies to b_1 - b_{n+1}. They are completely different techniques for different types of sequences.

Can telescoping series be infinite?

Yes, an infinite telescoping series can converge if the tail term approaches a limit. For example, \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right) = 1 because the partial sum simplifies to 1 - \frac{1}{n+1}, and \lim_{n \to \infty} \frac{1}{n+1} = 0.

Why is it important to expand the series and show cancellations explicitly?

Expanding lets you verify that the series really does telescope and see exactly which terms survive after cancellation. This process prevents mistakes like forgetting the negative sign on the last term or failing to recognize that the series can be rewritten into a telescoping form.

Telescoping Series

A telescoping series simplifies sums where consecutive terms partially cancel out, leaving only the first and last terms to compute.

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

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What each symbol means

What Telescoping Series takes

Symbol	Meaning
$b_k$	$b_k$ stands for the $k$ -th term of a general sequence that forms the building block of the series; if you read it as a single unknown value instead of a sequence, you lose the cancellation pattern that makes the method work.
$k$	$k$ is the index variable counting positions from 1 up to $n$ ; treating it as a constant instead of a placeholder for position would make you miss why consecutive terms cancel.
$n$	$n$ is the number of terms in the sum and the upper limit of the index; if you confuse it with a term in the sequence itself rather than a count of positions, you will not know where the telescoping stops.

When to use it

Use telescoping series when you can rewrite a sum as a difference of consecutive terms.

Level

Usually taught in: Calculus II

Worked examples

1. Telescoping fractions with small integers

Problem

Evaluate

\sum_{k=1}^{5} \left(\frac{1}{k} - \frac{1}{k+1}\right)

$\sum_{k=1}^{5} \left(\frac{1}{k} - \frac{1}{k+1}\right) = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{6}\right)$
Expand the sigma notation to write out all five terms.
$= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6}$
Rewrite the sum without parentheses to show exactly which terms appear positively and negatively.
$= 1 - \frac{1}{6} = \frac{5}{6}$
All middle terms cancel: $-\frac{1}{2} + \frac{1}{2} = 0$ , $-\frac{1}{3} + \frac{1}{3} = 0$ , and so on, leaving only $1$ and $-\frac{1}{6}$ .

Answer: $\frac{5}{6}$

This series telescopes because each negative fraction in one term equals the positive fraction at the start of the next term, causing cancellation and leaving only the first and last fractions.

2. Telescoping with products and negative results

Problem

Compute

\sum_{k=1}^{3} \left(k(k+1) - (k+1)(k+2)\right)

$\sum_{k=1}^{3} \left(k(k+1) - (k+1)(k+2)\right) = \left(1 \cdot 2 - 2 \cdot 3\right) + \left(2 \cdot 3 - 3 \cdot 4\right) + \left(3 \cdot 4 - 4 \cdot 5\right)$
Substitute $k = 1, 2, 3$ to expand each grouped term.
$= (2 - 6) + (6 - 12) + (12 - 20)$
Evaluate the products: $1 \cdot 2 = 2$ , $2 \cdot 3 = 6$ , $3 \cdot 4 = 12$ , $4 \cdot 5 = 20$ .
$= 2 - 6 + 6 - 12 + 12 - 20 = 2 - 20 = -18$
Rewrite without parentheses and cancel: $-6 + 6 = 0$ and $-12 + 12 = 0$ , leaving $2 - 20$ .

Answer: $-18$

The negative result shows that the last term dominates; the telescoping method still works even when the answer is negative, since only the first and last products in the sequence matter.

3. Real-world application: bake sale profit

Problem

The Math Club's daily profit for a bake sale is given by the formula

\frac{50}{k} - \frac{50}{k+1}

dollars on day

k

. What is their total profit for the first 10 days?

$\sum_{k=1}^{10} \left(\frac{50}{k} - \frac{50}{k+1}\right) = 50 \sum_{k=1}^{10} \left(\frac{1}{k} - \frac{1}{k+1}\right)$
Factor out the constant 50 from both fractions.
$= 50 \left[\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{10} - \frac{1}{11}\right)\right]$
Expand the telescoping series to show the first few terms and the last term; the middle terms are abbreviated with $\cdots$ but do cancel.
$= 50 \left(1 - \frac{1}{11}\right) = 50 \cdot \frac{10}{11} = \frac{500}{11}$
All middle fractions cancel because each $-\frac{1}{k}$ term matches a $+\frac{1}{k}$ term in the next pair; only $1$ and $-\frac{1}{11}$ remain.

Answer: $\frac{500}{11}$

This word problem demonstrates that telescoping is useful in real applications where a formula has an inherent telescoping structure; recognizing this pattern lets you compute a 10-term sum in just a few steps.

Common mistakes

Where Telescoping Series usually goes wrong

The mistake	Why it is wrong	The fix
Students see $b_1 - b_{n+1}$ and write $b_1 + b_{n+1}$ or forget the negative sign on $b_{n+1}$ .	The formula explicitly has a minus sign between $b_1$ and $b_{n+1}$ ; misreading it as addition or dropping the minus will give you the wrong answer with the wrong sign.	Always write out the formula $b_1 - b_{n+1}$ in full and double-check the sign before substituting numbers.
After expanding a telescoping series, students try to add all remaining terms together instead of recognizing the cancellation pattern.	If you do not deliberately pair up and cancel the terms, you will attempt tedious arithmetic on numbers that should eliminate each other, wasting time and risking computational error.	After expanding, rearrange the sum so that $-b_2 + b_2$ , $-b_3 + b_3$ , and so on are clearly adjacent, then cross them out as 0 before you do any arithmetic.
When given a series like $\sum_{k=1}^{n} \frac{1}{k(k+1)}$ , students do not recognize that it can be rewritten as a telescoping series and instead try to sum it term-by-term.	Not all series are already written as explicit differences; you must recognize when algebraic decomposition can split a series into a telescoping form, or you will miss the shortcut entirely.	Before jumping to direct summation, check whether the series can be decomposed using the identity $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$ to reveal the telescoping structure.

Tips and when to use something else

Write out the first few and last few terms when you expand a telescoping series — this makes the cancellation pattern obvious and helps you avoid algebra mistakes.
If your series does not obviously telescope, look for opportunities to use algebraic identities or decomposition to rewrite it into a telescoping form.
Telescoping works wonderfully for finite sums, but if you extend to an infinite series, you still must verify that the tail term converges using the Ratio Test or Integral Test.
If a series is not telescoping and does not have a constant ratio or constant difference between terms, check whether it is a Geometric Series or Arithmetic Series instead before applying convergence tests.

Related formulas

Sigma NotationSigma notation compresses long sums into a compact formula, letting you write and calculate the sum of many terms efficiently and accurately.Arithmetic SeriesFind the sum of an arithmetic sequence without adding every term—just use the first term, last term, and how many terms there are.Infinite Geometric SeriesInfinite Geometric Series gives you the sum of all terms in a geometric sequence when the common ratio lies strictly between negative one and one.Ratio TestDetermine whether an infinite series converges by checking if the ratio of consecutive terms approaches a value less than one.

Frequently asked questions

How do I know if a series will telescope?: A series telescopes if you can rewrite it as a difference of consecutive terms from a sequence, where each term appears positively in one place and negatively in the next. If the series is not already in this form, try using algebraic identities or decomposition to convert it into telescoping form.
What's the difference between a telescoping series and a geometric series?: A geometric series has a constant ratio between consecutive terms and is summed using $S = \frac{a(1-r^n)}{1-r}$ . A telescoping series cancels out middle terms and simplifies to $b_1 - b_{n+1}$ . They are completely different techniques for different types of sequences.
Can telescoping series be infinite?: Yes, an infinite telescoping series can converge if the tail term approaches a limit. For example, $\sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right) = 1$ because the partial sum simplifies to $1 - \frac{1}{n+1}$ , and $\lim_{n \to \infty} \frac{1}{n+1} = 0$ .
Why is it important to expand the series and show cancellations explicitly?: Expanding lets you verify that the series really does telescope and see exactly which terms survive after cancellation. This process prevents mistakes like forgetting the negative sign on the last term or failing to recognize that the series can be rewritten into a telescoping form.

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Reviewed 2026-09-18