Telescoping Series

Learning Outcomes

Evaluate a telescoping series

Consider the series $\displaystyle\sum _{n=1}^{\infty }\frac{1}{n\left(n+1\right)}$ . We discussed this series in the example: Evaluating Limits of Sequences of Partial Sums, showing that the series converges by writing out the first several partial sums ${S}_{1},{S}_{2}\text{,}\ldots,{S}_{6}$ and noticing that they are all of the form ${S}_{k}=\frac{k}{k+1}$ . Here we use a different technique to show that this series converges. By using partial fractions, we can write

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

$\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}$ .

Therefore, the series can be written as

\sum_{n = 1}^{\infty} [\frac{1}{n} - \frac{1}{n + 1}] = (1 + \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots

$\displaystyle\sum _{n=1}^{\infty }\left[\frac{1}{n}-\frac{1}{n+1}\right]=\left(1+\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots$ .

Writing out the first several terms in the sequence of partial sums $\left\{{S}_{k}\right\}$ , we see that

\begin{array}{l} S_{1} = 1 - \frac{1}{2} \\ S_{2} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) = 1 - \frac{1}{3} \\ S_{3} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) = 1 - \frac{1}{4} . \end{array}

$\begin{array}{l}{S}_{1}=1-\frac{1}{2}\hfill \\ {S}_{2}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)=1-\frac{1}{3}\hfill \\ {S}_{3}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)=1-\frac{1}{4}.\hfill \end{array}$

In general,

S_{k} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{k} - \frac{1}{k + 1}) = 1 - \frac{1}{k + 1}

${S}_{k}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots +\left(\frac{1}{k}-\frac{1}{k+1}\right)=1-\frac{1}{k+1}$ .

We notice that the middle terms cancel each other out, leaving only the first and last terms. In a sense, the series collapses like a spyglass with tubes that disappear into each other to shorten the telescope. For this reason, we call a series that has this property a telescoping series. For this series, since ${S}_{k}=1 - \frac{1}{\left(k+1\right)}$ and $\frac{1}{\left(k+1\right)}\to 0$ as $k\to \infty$ , the sequence of partial sums converges to $1$ , and therefore the series converges to $1$ .

Definition

A telescoping series is a series in which most of the terms cancel in each of the partial sums, leaving only some of the first terms and some of the last terms.

For example, any series of the form

\sum_{n = 1}^{\infty} [b_{n} - b_{n + 1}] = (b_{1} - b_{2}) + (b_{2} - b_{3}) + (b_{3} - b_{4}) + \dots

$\displaystyle\sum _{n=1}^{\infty }\left[{b}_{n}-{b}_{n+1}\right]=\left({b}_{1}-{b}_{2}\right)+\left({b}_{2}-{b}_{3}\right)+\left({b}_{3}-{b}_{4}\right)+\cdots$

is a telescoping series. We can see this by writing out some of the partial sums. In particular, we see that

\begin{array}{l} S_{1} = b_{1} - b_{2} \\ S_{2} = (b_{1} - b_{2}) + (b_{2} - b_{3}) = b_{1} - b_{3} \\ S_{3} = (b_{1} - b_{2}) + (b_{2} - b_{3}) + (b_{3} - b_{4}) = b_{1} - b_{4} . \end{array}

$\begin{array}{l}{S}_{1}={b}_{1}-{b}_{2}\hfill \\ {S}_{2}=\left({b}_{1}-{b}_{2}\right)+\left({b}_{2}-{b}_{3}\right)={b}_{1}-{b}_{3}\hfill \\ {S}_{3}=\left({b}_{1}-{b}_{2}\right)+\left({b}_{2}-{b}_{3}\right)+\left({b}_{3}-{b}_{4}\right)={b}_{1}-{b}_{4}.\hfill \end{array}$

In general, the kth partial sum of this series is

S_{k} = b_{1} - b_{k + 1}

${S}_{k}={b}_{1}-{b}_{k+1}$ .

Since the kth partial sum can be simplified to the difference of these two terms, the sequence of partial sums $\left\{{S}_{k}\right\}$ will converge if and only if the sequence $\left\{{b}_{k+1}\right\}$ converges. Moreover, if the sequence ${b}_{k+1}$ converges to some finite number $B$ , then the sequence of partial sums converges to ${b}_{1}-B$ , and therefore

\sum_{n = 1}^{\infty} [b_{n} - b_{n + 1}] = b_{1} - B

$\displaystyle\sum _{n=1}^{\infty }\left[{b}_{n}-{b}_{n+1}\right]={b}_{1}-B$ .

In the next example, we show how to use these ideas to analyze a telescoping series of this form.

Example: Evaluating a Telescoping Series

Determine whether the telescoping series

\sum_{n = 1}^{\infty} [\cos (\frac{1}{n}) - \cos (\frac{1}{n + 1})]

$\displaystyle\sum _{n=1}^{\infty }\left[\cos\left(\frac{1}{n}\right)-\cos\left(\frac{1}{n+1}\right)\right]$

converges or diverges. If it converges, find its sum.

Show Solution

By writing out terms in the sequence of partial sums, we can see that

\begin{array}{ccc} S_{1} & = & \cos (1) - \cos (\frac{1}{2}) \\ S_{2} & = & (\cos (1) - \cos (\frac{1}{2})) + (\cos (\frac{1}{2}) - \cos (\frac{1}{3})) = \cos (1) - \cos (\frac{1}{3}) \\ S_{3} & = & (\cos (1) - \cos (\frac{1}{2})) + (\cos (\frac{1}{2}) - \cos (\frac{1}{3})) + (\cos (\frac{1}{3}) - \cos (\frac{1}{4})) \\ = & \cos (1) - \cos (\frac{1}{4}) . \end{array}

$\begin{array}{ccc}\hfill {S}_{1}& =\hfill & \cos\left(1\right)-\cos\left(\frac{1}{2}\right)\hfill \\ \hfill {S}_{2}& =\hfill & \left(\cos\left(1\right)-\cos\left(\frac{1}{2}\right)\right)+\left(\cos\left(\frac{1}{2}\right)-\cos\left(\frac{1}{3}\right)\right)=\cos\left(1\right)-\cos\left(\frac{1}{3}\right)\hfill \\ \hfill {S}_{3}& =\hfill & \left(\cos\left(1\right)-\cos\left(\frac{1}{2}\right)\right)+\left(\cos\left(\frac{1}{2}\right)-\cos\left(\frac{1}{3}\right)\right)+\left(\cos\left(\frac{1}{3}\right)-\cos\left(\frac{1}{4}\right)\right)\hfill \\ & =\hfill & \cos\left(1\right)-\cos\left(\frac{1}{4}\right).\hfill \end{array}$

In general,

S_{k} = \cos (1) - \cos (\frac{1}{k + 1})

${S}_{k}=\cos\left(1\right)-\cos\left(\frac{1}{k+1}\right)$ .

Since $\frac{1}{\left(k+1\right)}\to 0$ as $k\to \infty$ and $\cos{x}$ is a continuous function, $\cos\left(\frac{1}{\left(k+1\right)}\right)\to \cos\left(0\right)=1$ . Therefore, we conclude that ${S}_{k}\to \cos\left(1\right)-1$ . The telescoping series converges and the sum is given by

\sum_{n = 1}^{\infty} [\cos (\frac{1}{n}) - \cos (\frac{1}{n + 1})] = \cos (1) - 1

$\displaystyle\sum _{n=1}^{\infty }\left[\cos\left(\frac{1}{n}\right)-\cos\left(\frac{1}{n+1}\right)\right]=\cos\left(1\right)-1$ .

try it

Determine whether $\displaystyle\sum _{n=1}^{\infty }\left[{e}^{\frac{1}{n}}-{e}^{\frac{1}{\left(n+1\right)}}\right]$ converges or diverges. If it converges, find its sum.

Hint

Show Solution

$e - 1$

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Try It

Activity: Euler’s Constant

We have shown that the harmonic series $\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}$ diverges. Here we investigate the behavior of the partial sums ${S}_{k}$ as $k\to \infty$ . In particular, we show that they behave like the natural logarithm function by showing that there exists a constant $\gamma$ such that

\sum_{n = 1}^{k} \frac{1}{n} - ln k \to γ as k \to \infty

$\displaystyle\sum _{n=1}^{k}\frac{1}{n}-\text{ln}k\to \gamma \text{as}k\to \infty$ .

This constant $\gamma$ is known as Euler’s constant.

Let ${T}_{k}=\displaystyle\sum _{n=1}^{k}\frac{1}{n}-\text{ln}k$ . Evaluate ${T}_{k}$ for various values of $k$ .
For $T_{k}$ as defined in part 1. show that the sequence ${T_{k}}$ converges by using the following steps.
1. Show that the sequence $\left\{{T}_{k}\right\}$ is monotone decreasing. (Hint: Show that $\text{ln}\left(1+\frac{1}{k}>\frac{1}{\left(k+1\right)}\right)$
2. Show that the sequence $\left\{{T}_{k}\right\}$ is bounded below by zero. (Hint: Express $\text{ln}k$ as a definite integral.)
3. Use the Monotone Convergence Theorem to conclude that the sequence $\left\{{T}_{k}\right\}$ converges. The limit $\gamma$ is Euler’s constant.
Now estimate how far $T_{k}$ is from $γ$ for a given integer $k$ . Prove that for $k \geq 1$ , $0 < T_{k} - γ \leq \frac{1}{k}$ by using the following steps.
1. Show that $\text{ln}\left(k+1\right)-\text{ln}k<\frac{1}{k}$ .
2. Use the result from part a. to show that for any integer $k$ ,
  
  ${T}_{k}-{T}_{k+1}<\frac{1}{k}-\frac{1}{k+1}$ .
3. For any integers $k$ and $j$ such that $j>k$ , express ${T}_{k}-{T}_{j}$ as a telescoping sum by writing
  
  ${T}_{k}-{T}_{j}=\left({T}_{k}-{T}_{k+1}\right)+\left({T}_{k+1}-{T}_{k+2}\right)+\left({T}_{k+2}-{T}_{k+3}\right)+\cdots +\left({T}_{j - 1}-{T}_{j}\right)$ .
  
  Use the result from part b. combined with this telescoping sum to conclude that
  
  ${T}_{k}-{T}_{j}<\frac{1}{k}-\frac{1}{j}$ .
4. Apply the limit to both sides of the inequality in part c. to conclude that
  
  ${T}_{k}-\gamma \le \frac{1}{k}$ .
5. Estimate $\gamma$ to an accuracy of within $0.001$ .

Module 5: Sequences and Series