3.3 Telescoping Series

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We determine the convergence or divergence of a Telescoping Series.

Telescoping Series

A telescoping series is a special type of series whose terms cancel each out in such a way that it is relatively easy to determine the exact value of its partial sums. In general, such a series is formed as a sum of differences:

$\begin{align*} \sum_{n=1}^\infty \left(a_n -a_{n+1}\right) &= \lim_{N \to \infty} \sum_{n=1}^N \left(a_n -a_{n+1}\right)\\ &=\lim_{N \to \infty}\left[\left(a_1 - a_2\right) + \left(a_2 - a_3\right) + \cdots + \left(a_N - a_{N+1}\right) \right]\\ &= \lim_{N \to \infty}\left[a_1 - a_{N+1}\right] \end{align*}$

In practice, creating the telescoping effect frequently involves a partial fraction decomposition.

example 1 Consider the series $\sum _{n=1}^\infty \frac {1}{n^2 +n}$

The $N^{th}$ partial sum is given by

$S_N = \sum _{n=1}^N \frac {1}{n^2 +n} = \frac {1}{2} + \frac {1}{6} + \frac {1}{12} + \cdots + \frac {1}{N^2 - N} + \frac {1}{N^2 + N}$ Note that the second to last term above comes from letting $n = N-1$ , so that in the denominator we have $n^2 + n = (N-1)^2 + (N-1) = N^2 - 2N + 1 + N - 1 = N^2 -N.$ There does not appear to be an easier way to express this partial sum. However, if we rewrite $\frac {1}{n^2 +n}$ using a partial fraction decomposition, we will see a nice formula for the partial sum. The decomposition is: $\frac {1}{n^2 + n} = \frac {1}{n(n+1)} = \frac {A}{n} + \frac {B}{n+1},$ with $A = 1$ and $B=-1$ (verify). Now rewrite the $N^{th}$ partial sum using this decomposition:

$\begin{align*} S_N = \sum_{n=1}^N \frac{1}{n^2 +n} =&\sum_{n=1}^N \left(\frac{1}{n}-\frac{1}{n+1}\right)\\ =& \left(1 - \frac12 \right) + \left(\frac12 - \frac13 \right) + \left(\frac13 - \frac14 \right) + \cdots \\ &+ \left(\frac{1}{N-1} - \frac{1}{N} \right) +\left(\frac{1}{N} - \frac{1}{N+1} \right) \end{align*}$

Looking carefully at the terms in this sum, we see a lot of cancellation. The terms $-\frac 12$ and $+\frac 12$ cancel as do the pair $-\frac 13$ and $+\frac 13$ and the pair $-\frac {1}{N}$ and $+\frac {1}{N}$ . The only fractions that don’t cancel are the first, $1$ , and the last $-1/(N+1)$ . The series collapses like an old-time telescope and the $N^{th}$ partial sum is: $S_N = \sum _{n=1}^N \frac {1}{n^2 +n} = 1 - \frac {1}{N+1}.$ We can now find the sum of the series by taking a limit: $S = \sum _{n=1}^\infty \frac {1}{n^2 +n} = \lim _{N\to \infty } S_N = \lim _{N\to \infty } \left (1 - \frac {1}{N+1}\right ) = 1.$ Thus, the series converges and its sum is 1.

example 2 Consider the series $\sum _{n=1}^\infty \frac {1}{n^2 +2n}$

We begin with a partial fraction decomposition: $\frac {1}{n^2 + 2n} = \frac {1}{n(n+2)} = \frac {A}{n} + \frac {B}{n+2}$ with $A = 1/2$ and $B=-1/2$ (verify). Thus, $\sum _{n=1}^\infty \frac {1}{n^2 + 2n} = \frac 12 \sum _{n=1}^\infty \left ( \frac {1}{n} - \frac {1}{n+2} \right )$ Using the telescoping form, we can find $S_N$ :

$\begin{align*} \frac12 \sum_{n=1}^N \left(\frac{1}{n}-\frac{1}{n+2}\right) &= \frac12 \Bigg[\left(1 - \frac13 \right) + \left(\frac12 - \frac14 \right) + \left(\frac13 - \frac15 \right) + \cdots \\ & + \left(\frac{1}{N-2} - \frac{1}{N} \right) +\left(\frac{1}{N-1} - \frac{1}{N+1} \right) +\left(\frac{1}{N} - \frac{1}{N+2} \right) \Bigg] \end{align*}$

Observe that the first negative term is $-1/3$ and the last positive term is $1/N$ . This means that the first two positive terms, $1$ and $1/2$ and the last two negative terms, $-1/(N+1)$ and $-1/(N+2)$ will survive the cancellation: $S_N = \frac 12 \left [ 1 + \frac 12 - \frac {1}{N+1} - \frac {1}{N+2}\right ]$ We can now find the sum of the series as a limit of its partial sums: $S = \lim _{N\to \infty } S_N = \frac 12 \left [1 + \frac 12 - 0- 0\right ] = \frac 34$ Thus, the series converges and its sum is 3/4.

(problem 1) Find a formula for $S_N$ , and find the sum of the series (if it converges) or state that it diverges: $\sum _{n=0}^\infty \frac {1}{n^2 + 3n + 2}$ The partial fraction decomposition has the form $\frac {A}{n+1} + \frac {B}{n+2}$ with $A = \answer {1} \quad \mbox {and} \quad B = \answer {-1}$ The $N^{th}$ partial sum is $\displaystyle S_N = \sum _{n=0}^N \frac {1}{n^2 + 3n + 2} = \answer {1 - 1/(N+2)}$
The limit of the partial sums is $\displaystyle S = \lim _{N\to \infty }S_N = \answer {1}$
The series convergesdiverges to $\answer {1}$

(problem 2) Find a formula for $S_N$ , and find the sum of the series (if it converges) or state that it diverges: $\sum _{n=0}^\infty \frac {1}{n^2 + 4n + 3}$

The partial fraction decomposition has the form $\frac {A}{n+1} + \frac {B}{n+3}$ with $A = \answer {1/2} \quad \mbox {and} \quad B = \answer {-1/2}$ The $N^{th}$ partial sum is $\displaystyle S_N = \sum _{n=0}^N \frac {1}{n^2 + 4n + 3} = \answer {3/4 - 1/(2N+4)- 1/(2N+6)}$
The limit of the partial sums is $\displaystyle S = \lim _{N\to \infty }S_N = \answer {3/4}$
The series to $\answer {3/4}$

(problem 3) Find a formula for $S_N$ , and find the sum of the series (if it converges) or state that it diverges: $\sum _{n=1}^\infty \left ( 2^{\frac {1}{n}} - 2^{\frac {1}{n+1}}\right )$ The $N^{th}$ partial sum is $\displaystyle S_N = \sum _{n=1}^N \left ( 2^{\frac {1}{n}} - 2^{\frac {1}{n+1}}\right ) = \answer {2 - 2^{1/(N+1)}}$
The limit of the partial sums is $\displaystyle S = \lim _{N\to \infty }S_N = \answer {1}$
The series convergesdiverges to $\answer {1}$

(problem 4) Find a formula for $S_N$ , and find the sum of the series (if it converges) or state that it diverges: $\sum _{n=2}^\infty \ln \left (1 + \frac {1}{n}\right )$

Writing $1 + \frac {1}{n}$ as a fraction gives

$\frac {2}{n}$
$\frac {2}{n+1}$
$\frac {n+1}{n}$

Next, we use the log property $\ln \frac {u}{v} = \ln u - \ln v$ to write $\ln \left (1 + \frac {1}{n}\right )$ as

$\ln n - \ln (n+1)$
$\ln (n+1) - \ln n$
$1 - \ln n$

The $N^{th}$ partial sum is $\displaystyle S_N = \sum _{n=2}^N \ln \left (1 + \frac {1}{n}\right ) = \answer {\ln (N+1)}$

The limit of the partial sums is $\displaystyle S = \lim _{N\to \infty }S_N = \answer {\infty }$

The series convergesdiverges

(problem 5) Find a formula for $S_N$ , and find the sum of the series (if it converges) or state that it diverges: $\sum _{n=1}^\infty \frac {1}{\sqrt n + \sqrt {n+1}}$ Multiplying by the conjugate radical, $\sqrt {n+1} - \sqrt {n}$ , over itself yields $\frac {1}{\sqrt n + \sqrt {n+1}} \cdot \frac {\sqrt {n+1} - \sqrt {n}}{\sqrt {n+1} - \sqrt {n}} = \answer {\sqrt {n+1} -\sqrt {n}}$

The $N^{th}$ partial sum is $\displaystyle S_N = \sum _{n=1}^N \frac {1}{\sqrt n + \sqrt {n+1}} = \answer {\sqrt {N+1} - 1}$

The limit of the partial sums is $\displaystyle S = \lim _{N\to \infty }S_N = \answer {\infty }$

The series convergesdiverges

(problem 6) Find a formula for $S_N$ , and find the sum of the series (if it converges) or state that it diverges: $\sum _{n=0}^\infty (-1)^n$

If $N$ is even, then the $N^{th}$ partial sum is $\displaystyle S_N = \sum _{n=0}^N (-1)^n = \answer {1}$

If $N$ is odd, then the $N^{th}$ partial sum is $\displaystyle S_N = \sum _{n=0}^N (-1)^n = \answer {0}$

The limit of the partial sums is $\displaystyle S = \lim _{N\to \infty }S_N = \answer {DNE}$

The series convergesdiverges