Telescoping Series
Textbook Reference
| Primary source | Stewart, Calculus, 9th edition, Section 11.2: “Series,” pages 778 to 779 (Example 2, the telescoping sum) and Exercises 17 to 22 |
| Open companion | OpenStax Calculus Volume 2, Section 5.2: “Infinite Series” |
| Companion link | https://openstax.org/books/calculus-volume-2/pages/5-2-infinite-series |
Telescoping series build on the partial-sum definition of a series. They are the second family (after geometric series) whose sum can be found exactly.
Key idea
A telescoping series is one where the inside of the sum cancels itself, leaving only the two ends.
The name comes from an old collapsing spyglass: the long tube folds down into a short stack. A telescoping series does the same thing. If you can write each term as a difference of two pieces, $b_n - b_{n+1}$, then in the partial sum the second piece of one term cancels the first piece of the next: \[ (b_1 - b_2) + (b_2 - b_3) + (b_3 - b_4) + \cdots + (b_n - b_{n+1}). \] Every interior $b$ appears once with a plus and once with a minus, so it vanishes. All that survives is the very first piece and the very last piece: $s_n = b_1 - b_{n+1}$.
That cancellation is the technique, and it is useful because it gives you a closed form for the partial sum, which is exactly what you need to take the limit and find the exact sum. The work is almost entirely in the first step: rewriting each term as a difference, usually with partial fractions. Once the difference is in hand, the cancellation does the rest.
Prerequisite Check
If these are solid, you are ready.
Quick Reference
Telescoping series. A series $\displaystyle\sum a_n$ is telescoping if each term can be written as a difference \[ a_n = b_n - b_{n+1} \] so that consecutive terms cancel in the partial sum.
The collapsed partial sum. When $a_n = b_n - b_{n+1}$, \[ s_n = \sum_{i=1}^{n}(b_i - b_{i+1}) = b_1 - b_{n+1}. \]
The sum. \[ \sum_{n=1}^{\infty} a_n = \lim_{n\to\infty} s_n = b_1 - \lim_{n\to\infty} b_{n+1}, \] which is a finite number (so the series converges) exactly when $\lim_{n\to\infty} b_{n+1}$ exists.
Procedure.
- Write the general term as a difference $b_n - b_{n+1}$, usually by partial fractions.
- Write out the first few terms of the partial sum and confirm the cancellation pattern.
- Identify the surviving end pieces to get $s_n$ in closed form.
- Take $\displaystyle\lim_{n\to\infty} s_n$ to get the sum.
Caution. Always confirm which end pieces survive by writing out the first two and last two terms. Some series cancel with a gap (leaving more than one term on each end).
Key Concepts
1. The Cancellation, Made Explicit
The reason a telescoping series can be summed exactly is that its partial sum has a closed form, just as a geometric series does. The mechanism is cancellation rather than algebra. Suppose $a_n = b_n - b_{n+1}$. Then \[ s_n = (b_1 - b_2) + (b_2 - b_3) + (b_3 - b_4) + \cdots + (b_{n-1} - b_n) + (b_n - b_{n+1}). \] Read down the list: the $-b_2$ in the first group cancels the $+b_2$ in the second, the $-b_3$ cancels the $+b_3$, and so on. Only the opening $b_1$ and the closing $-b_{n+1}$ are left: \[ s_n = b_1 - b_{n+1}. \] Now the sum is a single limit.
2. The Standard Worked Example
Example 1 (the model telescoping series). Show that $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ converges, and find its sum.
Goal. Split the term by partial fractions into a difference, watch the partial sum collapse, then take the limit.
Work. Partial fractions give \[ \frac{1}{i(i+1)} = \frac{1}{i} - \frac{1}{i+1}, \] so $b_i = \dfrac{1}{i}$. The partial sum is \[ s_n = \sum_{i=1}^{n}\left(\frac{1}{i} - \frac{1}{i+1}\right) = \left(1 - \tfrac{1}{2}\right) + \left(\tfrac{1}{2} - \tfrac{1}{3}\right) + \cdots + \left(\tfrac{1}{n} - \tfrac{1}{n+1}\right). \] Everything in the middle cancels, leaving the first piece $1$ and the last piece $-\dfrac{1}{n+1}$: \[ s_n = 1 - \frac{1}{n+1}. \] Then \[ \lim_{n\to\infty} s_n = 1 - 0 = 1. \]
Answer: the series converges and $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+1)} = 1$.
Recap. The two surviving terms are the opening $b_1 = 1$ and the vanishing tail $b_{n+1} = \dfrac{1}{n+1}$. The sum is $b_1$ minus the limit of the tail, here $1 - 0 = 1$.
(Common error: writing $s_n = 0$ because “everything cancels.” The interior cancels; the two end pieces do not. Always keep $b_1$ and $-b_{n+1}$.)
3. When the Cancellation Leaves a Gap
Not every telescoping series leaves exactly one term on each end. When the difference skips an index, more than one term survives at each end.
Example 2 (a wider gap). Find the sum of $\displaystyle\sum_{n=1}^{\infty} \frac{3}{n(n+3)}$.
Goal. Decompose, then track carefully which terms survive, because the gap is three wide.
Work. Partial fractions give \[ \frac{3}{n(n+3)} = \frac{1}{n} - \frac{1}{n+3}, \] so each term subtracts a piece three positions ahead. Writing the partial sum, \[ s_n = \left(1 - \tfrac{1}{4}\right) + \left(\tfrac{1}{2} - \tfrac{1}{5}\right) + \left(\tfrac{1}{3} - \tfrac{1}{6}\right) + \left(\tfrac{1}{4} - \tfrac{1}{7}\right) + \cdots \] The $-\tfrac{1}{4}$ in the first group cancels the $+\tfrac{1}{4}$ in the fourth group, and so on. The first three positive pieces ($1, \tfrac{1}{2}, \tfrac{1}{3}$) never get canceled, and the last three negative pieces survive at the tail: \[ s_n = \left(1 + \frac{1}{2} + \frac{1}{3}\right) - \left(\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3}\right). \] As $n\to\infty$ the three tail terms each go to $0$, so \[ \sum_{n=1}^{\infty}\frac{3}{n(n+3)} = 1 + \frac{1}{2} + \frac{1}{3} = \frac{11}{6}. \]
Answer: $\dfrac{11}{6}$
Recap. When the difference skips $k$ positions, $k$ terms survive at the front and $k$ at the back. Writing out the first several groups is the only reliable way to see exactly which terms remain.
(Common error: assuming only one term survives at each end. A term of the form $\dfrac{1}{n} - \dfrac{1}{n+3}$ leaves three uncanceled pieces on each side. Do not pattern-match to the $1/(n(n+1))$ case without checking the gap.)
Inline Self-Check
Question. Use the decomposition to find $\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{n} - \frac{1}{n+1}\right)$ directly.
Show answer
This is already a difference with $b_n = \dfrac{1}{n}$. The partial sum collapses to $s_n = 1 - \dfrac{1}{n+1}$, and the limit is $1 - 0 = 1$. The sum is $1$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Concluding $s_n = 0$ | “all terms cancel” | The interior cancels; keep $b_1$ and $-b_{n+1}$ |
| Missing surviving terms with a gap | Treating $\frac{1}{n}-\frac{1}{n+3}$ like a one-step difference | A $k$-step gap leaves $k$ terms on each end |
| Wrong partial fractions | $\frac{1}{n(n+1)} = \frac{1}{n} + \frac{1}{n+1}$ | The decomposition is a difference, $\frac{1}{n} - \frac{1}{n+1}$ |
| Forgetting the limit of the tail | Reporting $s_n$ as the sum | The sum is $\lim_{n\to\infty} s_n$, after the tail goes to $0$ |
| Not writing out enough terms | Guessing the pattern | Write the first two and last two groups to see the cancellation |
Common Misconceptions
in a telescoping series all terms cancel, so the sum is zero.
This is the concept-image-conflicts-definition error about what cancels. Interior terms cancel in pairs, but the first piece and the last tail piece survive. For $\sum (1/n - 1/(n+1))$, the partial sum collapses to $s_n = 1 - 1/(n+1)$, not to $0$. The sum is $\lim s_n = 1$. Concluding that everything cancels ignores the boundary terms that have no matching partner.
a gap of more than one position in the telescoping difference leaves only one term at each end.
This is the concept-image-conflicts-definition error caused by pattern-matching to the simplest case. When the term is $1/n - 1/(n+3)$, the cancellation skips three positions, leaving the first three positive terms and three final negative terms uncanceled. Only a one-step difference $b_n - b_{n+1}$ leaves exactly one term at each end. Always write out the first and last several groups to count the surviving pieces.
Leveled Practice
Attempt each problem before opening the answer.
Level 1: Direct Application
Problem 1. Find the sum of $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n(n-1)}$.
Show answer
Decompose: $\dfrac{1}{n(n-1)} = \dfrac{1}{n-1} - \dfrac{1}{n}$. Starting at $n = 2$, the partial sum is \[ \left(1 - \tfrac{1}{2}\right) + \left(\tfrac{1}{2} - \tfrac{1}{3}\right) + \cdots + \left(\tfrac{1}{n-1} - \tfrac{1}{n}\right) = 1 - \frac{1}{n}. \] The limit is $1$. The sum is $1$.
Problem 2. Determine whether $\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}\right)$ converges, and find the sum if it does.
Show answer
The term is already a difference with $b_n = \dfrac{1}{\sqrt{n}}$. The partial sum is $s_n = 1 - \dfrac{1}{\sqrt{n+1}}$, and the limit is $1 - 0 = 1$. The series converges to $1$.
Problem 3. Find the sum of $\displaystyle\sum_{n=1}^{\infty}\frac{1}{(n+2) (n+3)}$.
Show answer
Decompose: $\dfrac{1}{(n+2)(n+3)} = \dfrac{1}{n+2} - \dfrac{1}{n+3}$, so $b_n = \dfrac{1}{n+2}$. The partial sum is $s_n = \dfrac{1}{3} - \dfrac{1}{n+3}$ (the first surviving piece is $b_1 = \tfrac{1}{3}$). The limit is $\dfrac{1}{3}$.
Level 2: Decompose First
Problem 4. Find the sum of $\displaystyle\sum_{n=1}^{\infty}\frac{2}{n(n+2)}$.
Show answer
Decompose: $\dfrac{2}{n(n+2)} = \dfrac{1}{n} - \dfrac{1}{n+2}$, a two-step gap. The first two positive pieces survive at the front, and two negative pieces at the tail: \[ s_n = \left(1 + \frac{1}{2}\right) - \left(\frac{1}{n+1} + \frac{1}{n+2}\right). \] The limit is $1 + \dfrac{1}{2} = \dfrac{3}{2}$.
Problem 5. Determine whether $\displaystyle\sum_{n=1}^{\infty}\ln\frac{n}{n+1}$ converges (note $\ln\frac{n}{n+1} = \ln n - \ln(n+1)$), and find the sum if it does.
Show answer
With $b_n = \ln n$, the partial sum is \[ s_n = \ln 1 - \ln(n+1) = -\ln(n+1). \] As $n\to\infty$, $-\ln(n+1) \to -\infty$. The partial sums have no finite limit, so the series diverges.
(This is a telescoping series whose surviving tail does NOT go to a finite value, so it diverges.)
Level 3: Reasoning
Problem 6. A telescoping series has $a_n = b_n - b_{n+1}$. State the exact condition on $\{b_n\}$ under which the series converges, and give the sum in that case.
Show answer
The partial sum is $s_n = b_1 - b_{n+1}$. The series converges exactly when $\displaystyle\lim_{n\to\infty} b_{n+1}$ exists as a finite number $L$. In that case the sum is \[ \sum_{n=1}^{\infty} a_n = b_1 - L. \] If $\{b_n\}$ has no finite limit (as with $b_n = \ln n$), the series diverges.
Problem 7. Explain why a telescoping series and a geometric series are the only two families in this section whose sums can be found exactly, while later series require convergence tests instead.
Show answer
For both families the $n$th partial sum $s_n$ has a closed-form expression: $s_n = b_1 - b_{n+1}$ for telescoping, and $s_n = \dfrac{a(1 - r^n)}{1 - r}$ for geometric. Because the sum of a series is by definition $\lim_{n\to\infty} s_n$, having $s_n$ in closed form lets you compute the exact sum with a single limit. For a general series there is no such closed form for $s_n$, so the sum cannot be read off directly; the convergence tests later in the chapter decide whether the limit exists without ever producing its value.
Mastery Checklist
You have mastered this skill when you can do all of the following without notes:
Mental Model
Picture an old hand telescope made of nested tubes. Pull it open and it is long; collapse it and the inner tubes slide inside one another until only the two end caps remain visible. A telescoping series collapses the same way. Each term is a tube; the matching minus and plus pieces are the overlapping sections of neighboring tubes, and they slide into each other and disappear. When the dust settles, only the first cap ($b_1$) and the last cap ($-b_{n+1}$) are left.
The convergence question is then just: does the last cap settle to a fixed place as you slide it out to infinity? If $b_{n+1}$ approaches a finite value, the collapsed length settles and the series converges. If the last cap runs off to infinity (as $\ln(n+1)$ does), the series diverges.
Connections
Built From
- Infinite series and partial sums: the method finds a closed form for $s_n$ and takes its limit, which is the definition of a series sum.
- Partial fractions: the algebraic tool that turns a single fraction into the difference needed for cancellation.
Leads To
- The integral test and the comparison tests: the next sections handle series whose partial sums have no closed form. Telescoping and geometric series are the two cases where a closed form exists, so they are the baseline against which the harder cases are measured.
Why It Matters
Telescoping sums appear wherever a quantity is expressed as a running difference: the net change of a function over many small steps (the discrete cousin of the Fundamental Theorem of Calculus), the cancellation in a partial-fraction integration, and identities in combinatorics and probability where consecutive terms subtract cleanly. In computing, a telescoping sum is the closed form behind a prefix-sum or difference-array computation, where storing differences lets you reconstruct totals cheaply.
Audience Notes
For students who find math intimidating: the hard part is just the first step, rewriting the term as a difference. After that, you write out a few terms, cross off the middle, and keep the ends.
For students who want depth: a telescoping sum is the discrete analog of the Fundamental Theorem of Calculus: a sum of differences collapses to the endpoints, just as an integral of a derivative collapses to the boundary values. Seeing that parallel ties this technique to the rest of calculus.
For students aimed at a career in computing or engineering: the prefix-sum and difference-array patterns are telescoping sums in disguise. Storing $b_n$ and reading off differences is how range updates and range queries are made efficient.
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