Telescoping Sums with Multiple Residual Terms

Why More Than One Term Can Survive

A telescoping sum is a sum in which successive terms cancel after they are written in a suitable form. When each term has the form $b_k-b_{k+1}$, cancellation leaves one initial term and one final term. If the shift is larger, as in $b_k-b_{k+m}$ with $m>1$, cancellation still occurs, but it leaves several terms at each end.

Keeping track of all those residual terms is essential. Omitting even one changes the finite sum and can give an incorrect infinite limit. The method below gives a formula for any positive integer shift and shows exactly which terms remain.

Definition: A shifted difference with shift $m$ is a term of the form $b_k-b_{k+m}$, where $m$ is a fixed positive integer and $b_k$ is a sequence. A finite sum of these terms is telescoping because terms with matching indices cancel.

The Finite-Sum Formula

For a sum beginning at $k=1$, the first $N$ shifted differences satisfy the following formula. Here $N$ and $m$ are positive integers.

\sum_{k=1}^{N}(b_k-b_{k+m}) = \sum_{k=1}^{m}b_k-\sum_{k=N+1}^{N+m}b_k.

The first sum on the right contains the $m$ initial terms $b_1,\ldots,b_m$. The second contains the $m$ final terms $b_{N+1},\ldots,b_{N+m}$. These are the residual terms: the initial group has positive signs, and the final group has negative signs.

The formula also works when $N<m$. In that case, some indices occur in both groups on the right and cancel there, matching the cancellation in the original sum. The formula does not require the two groups to be disjoint.

Why the Formula Works

Separate the two sums and reindex the shifted one. In the second sum, set $j=k+m$, so its index runs from $m+1$ through $N+m$.

\begin{aligned} \sum_{k=1}^{N}(b_k-b_{k+m}) &=\sum_{k=1}^{N}b_k-\sum_{k=1}^{N}b_{k+m}\\ &=\sum_{k=1}^{N}b_k-\sum_{j=m+1}^{N+m}b_j. \end{aligned}

In the first sum, the terms with indices from $m+1$ through $N$, when present, cancel the terms with those same indices in the reindexed second sum. The indices left over at the beginning are $1,\ldots,m$; those left over at the end are $N+1,\ldots,N+m$, with negative signs. This gives the stated formula.

The same result can be written with any starting index $p$. For integers $p\leq N$, reindexing gives:

\sum_{k=p}^{N}(b_k-b_{k+m}) = \sum_{k=p}^{p+m-1}b_k-\sum_{k=N+1}^{N+m}b_k.

This version is useful when the finite sum does not begin at $k=1$. Its initial group always consists of the $m$ terms beginning at the actual starting index $p$.

Worked Example: A Shift of Two in a Reciprocal Sum

Evaluate the finite sum and its infinite limit for the terms $1/(k(k+2))$.

First write the summand as a shifted difference. Since $1/k-1/(k+2)=2/(k(k+2))$, we have:

\frac{1}{k(k+2)} =\frac12\left(\frac1k-\frac1{k+2}\right).

Apply the formula with $m=2$, including the factor $1/2$:

\begin{aligned} S_N &=\sum_{k=1}^{N}\frac{1}{k(k+2)}\\ &=\frac12\left(1+\frac12-\frac1{N+1}-\frac1{N+2}\right). \end{aligned}

For a finite check, take $N=3$. The formula gives:

S_3 =\frac12\left(1+\frac12-\frac14-\frac15\right) =\frac12\left(\frac{21}{20}\right) =\frac{21}{40}.

Directly adding the three terms gives the same result:

\frac1{1\cdot3}+\frac1{2\cdot4}+\frac1{3\cdot5} =\frac13+\frac18+\frac1{15} =\frac{40+15+8}{120} =\frac{63}{120} =\frac{21}{40}.

As $N$ increases, both $1/(N+1)$ and $1/(N+2)$ approach zero. Therefore the infinite sum, defined as the limit of its finite partial sums, is:

\lim_{N\to\infty}S_N =\frac12\left(1+\frac12\right) =\frac34.

The finite computation confirms the residual-term formula, and the limit follows by taking the limit of that formula.

From Finite Residuals to an Infinite Limit

An infinite series is evaluated through its partial sums: its sum is the limit of those partial sums if the limit exists. For a shifted difference, the finite formula makes the question precise. The limit depends on the final block of $m$ terms.

Theorem: Suppose $m$ is a fixed positive integer and $b_n\to L$ for a finite real number $L$. Then the partial sums of $\sum_{k=1}^{\infty}(b_k-b_{k+m})$ converge, and $$ \sum_{k=1}^{\infty}(b_k-b_{k+m}) = \sum_{k=1}^{m}b_k-mL. $$ In particular, if $b_n\to0$, the sum is $b_1+\cdots+b_m$.

To justify the result, apply the finite formula. For each of the fixed $m$ indices $N+1,\ldots,N+m$, the corresponding term $b_{N+r}$ approaches $L$ as $N\to\infty$. Thus the sum of the final $m$ terms approaches $mL$. The initial sum is fixed, so the partial sums approach $\sum_{k=1}^{m}b_k-mL$. When $L=0$, the final block vanishes in the limit.

This condition is a convenient way to evaluate many telescoping series; it is not a claim that $b_n\to0$ is necessary for every shifted series to converge. What the finite formula requires for convergence is that the final block have a limit. If $b_n$ itself has a finite limit, then the block limit is $mL$, as above.

Worked Example: A Shift of Three and Its Infinite Sum

Evaluate the partial sum and infinite sum of $\sum_{k=1}^{\infty}\left(\frac1{k+1}-\frac1{k+4}\right)$.

Set $b_k=1/(k+1)$ and $m=3$. The finite formula gives:

\begin{aligned} S_N &=\sum_{k=1}^{N}\left(\frac1{k+1}-\frac1{k+4}\right)\\ &=b_1+b_2+b_3-b_{N+1}-b_{N+2}-b_{N+3}\\ &=\frac12+\frac13+\frac14 -\frac1{N+2}-\frac1{N+3}-\frac1{N+4}. \end{aligned}

For $N=2$, this formula gives:

S_2 =\frac12+\frac13+\frac14-\frac14-\frac15-\frac16 =\frac{65-15-12-10}{60} =\frac{28}{60} =\frac7{15}.

Checking directly, the two summands are:

\left(\frac12-\frac15\right)+\left(\frac13-\frac16\right) =\frac3{10}+\frac16 =\frac9{30}+\frac5{30} =\frac7{15}.

Since $b_k=1/(k+1)\to0$, the final three terms tend to zero. The infinite sum is the sum of the first three terms of the sequence $b_k$:

\sum_{k=1}^{\infty}\left(\frac1{k+1}-\frac1{k+4}\right) =\frac12+\frac13+\frac14 =\frac{6+4+3}{12} =\frac{13}{12}.

The finite check agrees with direct addition, and the infinite value follows because each of the three residual terms tends to zero.

Starting at a Different Index

Changing the starting index changes the initial residual terms. It does not change the rule: retain the first $m$ terms starting at the actual lower index, then subtract the final $m$ terms after the upper index.

Worked Example: A Shift of Two with a Nonunit Starting Index

Evaluate $\sum_{k=3}^{6}(1/k-1/(k+2))$.

Here $b_k=1/k$, the shift is $m=2$, the starting index is $p=3$, and the upper index is $N=6$. The formula for a general starting index gives:

\begin{aligned} \sum_{k=3}^{6}\left(\frac1k-\frac1{k+2}\right) &=b_3+b_4-b_7-b_8\\ &=\frac13+\frac14-\frac17-\frac18\\ &=\frac{56+42-24-21}{168}\\ &=\frac{53}{168}. \end{aligned}

To check, add the original terms directly:

\begin{aligned} \left(\frac13-\frac15\right) +\left(\frac14-\frac16\right) +\left(\frac15-\frac17\right) +\left(\frac16-\frac18\right) &=\frac{112+70+48+35}{840}\\ &=\frac{265}{840}\\ &=\frac{53}{168}. \end{aligned}

The terms $1/5$ and $1/6$ cancel in the direct sum. The two initial terms $1/3,1/4$ and the two final terms $1/7,1/8$ remain, exactly as the formula predicts.

A Common Mistake: Keeping Only One Term at Each End

A frequent error is to treat a shift by $m$ as though it were a shift by one. For example, in a shift of three, it is not enough to retain $b_1$ and subtract only $b_{N+3}$. The two intermediate residual terms at each end matter.

Worked Example: Correcting an Incomplete Cancellation

Consider $\sum_{k=1}^{N}(1/k-1/(k+3))$. An incorrect calculation might write:

\sum_{k=1}^{N}\left(\frac1k-\frac1{k+3}\right) \overset{\text{wrong}}{=}1-\frac1{N+3}.

This drops the terms $1/2$ and $1/3$ from the initial group, and $1/(N+1)$ and $1/(N+2)$ from the final group. The correct formula is:

\sum_{k=1}^{N}\left(\frac1k-\frac1{k+3}\right) =1+\frac12+\frac13-\frac1{N+1}-\frac1{N+2}-\frac1{N+3}.

For $N=2$, the incorrect expression gives $1-1/5=4/5$. The correct expression gives:

1+\frac12+\frac13-\frac13-\frac14-\frac15 =\frac{110-20-15-12}{60} =\frac{63}{60} =\frac{21}{20}.

Direct addition confirms the correction:

\left(1-\frac14\right)+\left(\frac12-\frac15\right) =\frac34+\frac3{10} =\frac{15+6}{20} =\frac{21}{20}.

The practical safeguard is simple: write down all $m$ initial terms and all $m$ final terms before simplifying.

How to Apply the Method Reliably

When a summand is presented as a difference of shifted terms, organize the calculation before cancelling. Earlier material on partial fractions can help express rational summands in a telescoping form; once that form is established, the residual-term rule handles the sum.

Identify the shift.
Write the summand as $b_k-b_{k+m}$ and record the positive integer $m$.

Record the summation limits.
For a sum from $p$ to $N$, use the actual starting index $p$ and upper index $N$.

List both residual groups.
Keep $b_p,\ldots,b_{p+m-1}$, then subtract $b_{N+1},\ldots,b_{N+m}$.

Take a limit only after the finite formula is clear.
Evaluate the limit of the final block; if $b_n\to0$, that block tends to zero.

Telescoping is useful because it converts a long finite sum into a fixed number of boundary terms. For an infinite sum, it also separates the answer into an initial contribution and a tail whose limit can be examined directly. The size of the shift determines how many terms survive: a shift of $m$ leaves up to $m$ terms at each end, not just one.

Check Your Understanding

Use the finite residual-term formula in each question, and check any numerical partial sum by direct addition.

Write the residual-term formula for $\sum_{k=1}^{N}(b_k-b_{k+4})$.
Evaluate $\sum_{k=1}^{3}(1/k-1/(k+2))$.
If $b_n\to0$, what is the value of $\sum_{k=1}^{\infty}(b_k-b_{k+3})$, provided the series is interpreted through its partial-sum limit?
For $\sum_{k=5}^{N}(b_k-b_{k+2})$, which terms form the initial residual group?
If $b_n\to L$, where $L$ is finite, what limit does the final block $b_{N+1}+\cdots+b_{N+m}$ approach?