Solution: Telescoping series For the following telescoping

Chapter 11, Problem 53E

(choose chapter or problem)

QUESTION:

47-58. Telescoping series For the following telescoping series, find a formula for the nth term of the sequence of partial sums $\left\{S_{n}\right\}$. Then evaluate $\lim_{n\rightarrow\infty}\ S_n$, to obtain the value of the series or state that the series diverges.

$\sum_{k=1}^{\infty} \frac{1}{(k+p)(k+p+1)}$, where p is a positive integer

Questions & Answers

QUESTION:

$\sum_{k=1}^{\infty} \frac{1}{(k+p)(k+p+1)}$, where p is a positive integer

ANSWER:

Problem 53E

Telescoping series For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate to obtain the value of the series or stale that the series diverges.

, where p is a positive integer

Answer ;

Step 1 ;

The given Telescoping series is $\sum\limits_{k=1}^{\infty{}}$ $\frac{1}{(\ k+p)(k+p+1)}$ , where p is a positive integer

In this problem we have to find the formula for ${n}^{th}$ term ${S}_{n}$ in $\sum\limits_{k=1}^{\infty{}}$ $\frac{1}{(\ k+p)(k+p+1)}$ and then we have to evaluate $\lim\limits_{{n} \rightarrow {\infty{}}}{S}_{n}$ or we have state that the series diverges.

Consider $\sum\limits_{k=1}^{\infty{}}$ $\frac{1}{(\ k+p)(k+p+1)}$ = $\sum\limits_{k=1}^{\infty{}}($ $\frac{1}{\ k+p}$ - $\frac{1}{k+p+1}$ ) , since $\frac{1}{\ k+p}$ - $\frac{1}{k+p+1}$ = $\frac{(k+p+1)\ -\ (\ k+p)}{(\ k+p)(k+p+1)}$

= $\frac{1}{(\ k+p)(k+p+1)}$ .

$\sum\limits_{k=1}^{\infty{}}$ $\frac{1}{(\ k+p)(k+p+1)}$ = $\sum\limits_{k=1}^{\infty{}}($ $\frac{1}{\ k+p}$ - $\frac{1}{k+p+1}$ ) ………..(1)

Let us first find the ${n}^{th}$ term of the sequence of partial sums

${{S}_{n}}_{\ }$ = $\sum\limits_{k=1}^{n}($ $\frac{1}{\ k+p}$ - $\frac{1}{k+p+1}$ ) ……………(2)

Substitute values for $k=1,2,...,(n-1),n$ we get

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ {S}_{n}=[\frac{1}{\ 1+p}\ -\ \frac{1}{1+p+1}]+[\frac{1}{\ 2+p}\ -\ \frac{1}{2+p+1}]+[\frac{1}{\ 3+p}\ -\ \frac{1}{3+p+1}]+...\ +[\frac{1}{\ (n-1)+p}\ -\ \frac{1}{(n-1)+p+1})]$ +[ $\frac{1}{\ n+p}$ - $\frac{1}{n+p+1}$ ]