Telescoping Series

Cancellation in Partial Sums

The geometric-series formula came from simplifying finite partial sums before taking a limit. A different kind of simplification occurs when each term in a series is the difference of two nearby quantities: most terms cancel when the partial sum is expanded. This cancellation is the defining feature of a telescoping series.

Let $(b_n)$ be a sequence, and suppose the terms of a series are $a_n=b_n-b_{n+1}$. The first few partial sums have the form

\begin{aligned} S_1&=b_1-b_2,\\ S_2&=(b_1-b_2)+(b_2-b_3),\\ S_3&=(b_1-b_2)+(b_2-b_3)+(b_3-b_4). \end{aligned}

Each negative term after $-b_2$ cancels with the same quantity as a positive term in the next summand. Only the first and last boundary terms remain. The exact finite identity is the starting point for deciding whether the infinite series converges.

Definition: A series $\sum_{n=1}^{\infty}a_n$ is called a telescoping series when its terms can be written in a form such as $a_n=b_n-b_{n+1}$ for some sequence $(b_n)$. Its partial sums then simplify by cancellation. More generally, a series is telescoping if its terms are differences of a sequence with a fixed finite shift, as in $a_n=b_n-b_{n+k}$ for a fixed positive integer $k$.

The word “telescoping” describes the cancellation in the finite partial sums; it does not itself guarantee convergence. To find the sum, first derive the partial-sum formula with its correct endpoints. Then take the limit, if that limit exists.

The Finite Telescoping Identity

The cancellation pattern can be recorded for any consecutive block of terms. The lower endpoint matters: starting at $m$, rather than at $1$, changes the first boundary term.

Theorem (Finite Telescoping Identity): Let $(b_n)$ be a sequence, and let $1\leq m\leq N$. Then $$ \sum_{n=m}^{N}(b_n-b_{n+1})=b_m-b_{N+1}. $$

Proof. Expanding the finite sum gives

\begin{aligned} \sum_{n=m}^{N}(b_n-b_{n+1}) &=(b_m-b_{m+1})+(b_{m+1}-b_{m+2})+\cdots+(b_N-b_{N+1})\\ &=b_m-b_{N+1}. \end{aligned}

For every index strictly between $m$ and $N+1$, the term $b_j$ occurs once with a positive sign and once with a negative sign. Those two occurrences cancel. The term $b_m$ occurs only positively, and $b_{N+1}$ occurs only negatively. When $m=N$, the sum has just one term, and the identity reads $b_N-b_{N+1}=b_N-b_{N+1}$, so the endpoint case is included. $\square$

Taking $m=1$ and $a_n=b_n-b_{n+1}$, the $N$th partial sum is $S_N=b_1-b_{N+1}$. Thus the convergence question for this series becomes a question about the boundary sequence $b_{N+1}$.

Theorem (Convergence Criterion for a Telescoping Series): Suppose $a_n=b_n-b_{n+1}$ for every $n\geq1$. Then $\sum_{n=1}^{\infty}a_n$ converges if and only if the sequence $(b_n)$ converges to a finite real number $L$. In that case, $$ \sum_{n=1}^{\infty}a_n=b_1-L. $$

Proof. By the Finite Telescoping Identity, for every $N\geq1$,

S_N=\sum_{n=1}^{N}a_n=b_1-b_{N+1}.

If $b_n\to L$, then $b_{N+1}\to L$ as $N\to\infty$. Therefore $S_N\to b_1-L$, so the series converges and has the stated sum.

Conversely, suppose the series converges, so its partial sums satisfy $S_N\to S$ for some real number $S$. Rearranging the same finite identity gives $b_{N+1}=b_1-S_N$, and hence $b_{N+1}\to b_1-S$. This proves that the sequence $(b_n)$ converges: its tail, beginning with $b_2$, has the indicated limit, and adding the single first term does not affect convergence. If $L=b_1-S$, then $S=b_1-L$. This proves both directions and the sum formula. $\square$

The criterion is about the boundary sequence $b_{N+1}$, not about whether individual terms look small or whether some of the terms have opposite signs. If the boundary sequence has no finite limit, the partial sums have no finite limit either. In particular, the formula does not justify replacing an infinite series by a cancellation argument without first checking what remains at the endpoints.

Three Applications of the Boundary-Term Formula

Worked Example: A Positive Series with Sum One

For $n\geq1$, factor the denominator and decompose the term:

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

since the right-hand side is $\frac{(n+1)-n}{n(n+1)}=\frac{1}{n(n+1)}$. Thus $b_n=1/n$. The partial sum through $N$ is

\begin{aligned} S_N &=\sum_{n=1}^{N}\frac{1}{n(n+1)} =\sum_{n=1}^{N}\left(\frac1n-\frac{1}{n+1}\right)\\ &=1-\frac{1}{N+1}. \end{aligned}

Because $1/(N+1)\to0$, the partial sums tend to $1$. Therefore $\sum_{n=1}^{\infty}1/[n(n+1)]=1$. Each term is positive, and the partial sums remain below $1$, as the exact expression $1-1/(N+1)$ confirms.

Worked Example: A Two-Step Cancellation Pattern

Consider $\sum_{n=1}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+3}\right)$. The terms at the two ends are separated by two indices, so cancellation leaves two initial boundary terms and two final boundary terms. For $N\geq2$, expanding the finite sum gives

\begin{aligned} S_N &=\left(\frac12-\frac14\right) +\left(\frac13-\frac15\right) +\cdots +\left(\frac{1}{N+1}-\frac{1}{N+3}\right)\\ &=\frac12+\frac13-\frac{1}{N+2}-\frac{1}{N+3}. \end{aligned}

The endpoint terms are distinct in this expansion when $N\geq2$. For $N=1$, there is only one summand, so it is better to check separately:

S_1=\frac12-\frac14=\frac14, \qquad \frac12+\frac13-\frac13-\frac14=\frac14.

Consequently, the displayed formula for $S_N$ also holds when $N=1$, although the two middle terms then overlap and cancel. Since both $1/(N+2)$ and $1/(N+3)$ tend to zero, the series converges and

\sum_{n=1}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+3}\right) =\frac12+\frac13 =\frac56.

This example illustrates why a finite-sum formula should be checked at its smallest allowed index. A boundary-term expression can be correct even when the terms described as “remaining” coincide and cancel in a short partial sum.

Worked Example: A Telescoping Series That Diverges

For $n\geq1$, logarithm laws give $\log(n+1)-\log n=\log(1+1/n)$, which is positive. The partial sums telescope:

\begin{aligned} S_N &=\sum_{n=1}^{N}\bigl(\log(n+1)-\log n\bigr)\\ &=\log(N+1)-\log 1 =\log(N+1). \end{aligned}

Here $b_n=\log n$, and $b_{N+1}=\log(N+1)$ does not converge to a finite real number; it grows without bound. Equivalently, $S_N\to+\infty$, so this series diverges. The terms do tend to zero: $a_n=\log(1+1/n)\to\log(1)=0$, by continuity of the logarithm. Thus the necessary condition for series convergence is satisfied, but the series still diverges. The example shows why $a_n\to0$ alone cannot establish convergence.

How to Use Telescoping Safely

A reliable calculation has three stages: find a difference representation, compute a finite partial sum with its actual endpoints, and only then take a limit. For example, when the difference is $b_n-b_{n+1}$, the last negative term in $S_N$ is $-b_{N+1}$, not $-b_N$. The index $N+1$ appears because the $N$th summand already contains $b_{N+1}$.

Rewrite each term.
Express $a_n$ as $b_n-b_{n+1}$, or identify another fixed-shift difference that matches the terms.

Keep the sum finite.
Write out the terms in $S_N$ and cancel only occurrences that appear with opposite signs.

Check the endpoints.
Record exactly which terms remain, and separately check a small $N$ if the boundary terms can overlap.

Take the limit.
Use the remaining boundary terms to decide whether $S_N$ has a finite limit and, if so, compute the sum.

A common pitfall is to treat cancellation as if it were an operation on an infinite list of terms, without first identifying a sequence of finite partial sums. The definition of series convergence concerns the limit of those partial sums. The Finite Telescoping Identity supplies exact finite formulas, and the Convergence Criterion for a Telescoping Series then reduces the limit question to the boundary sequence.

Another pitfall is to assume that a familiar-looking difference must converge. In the first example, the boundary sequence $1/n$ tends to zero, leaving the finite initial value $b_1$. In the logarithmic example, the boundary sequence grows without bound, so the partial sums do too. Telescoping reveals the behavior clearly, but the boundary terms—not the cancellation alone—determine the outcome.

The fixed-shift example also hints at a broader pattern. When a term has the form $b_n-b_{n+k}$, cancellation can leave several terms at each end rather than just one. The same finite-sum discipline applies: identify the endpoints for a finite $N$, account for any overlap in short sums, and take limits only after obtaining a correct formula.

Takeaway: If $a_n=b_n-b_{n+1}$, then $S_N=b_1-b_{N+1}$. The series converges exactly when $b_n$ has a finite limit, and its sum is $b_1-\lim_{n\to\infty}b_n$.

Check Your Understanding

Use finite partial sums and their boundary terms to answer the following questions.

For $a_n=b_n-b_{n+1}$, what is the exact expression for $S_N$, and why does the final index equal $N+1$?
Find the sum of $\sum_{n=1}^{\infty}\left(\frac1n-\frac{1}{n+2}\right)$. Check the formula at $N=1$ before taking the limit.
Does $\sum_{n=1}^{\infty}\left(\frac{1}{\sqrt n}-\frac{1}{\sqrt{n+1}}\right)$ converge? If it does, find its sum.
For a telescoping series with $a_n=b_n-b_{n+1}$, what does convergence of the series imply about $(b_n)$?
Why does the fact that the terms in the logarithmic example tend to zero not imply that its series converges?

What You'll Learn