Telescoping series

**acevipa** · September 16th 2010, 05:52 AM

Can someone check my solution:

Consider the series $\sum_{k=1}^{\infty}a_k$ where $a_k=<br /> \frac{1}{k(k+1)}$

1) Find the partial fractions decomposition of $a_k$

$a_k=\frac{1}{k}-\frac{1}{k+1}$

2) Find a simple formula for the partial sum $s_n$ , where $s_n=\sum_{k=1}^{n}a_k$

$s_n=\sum_{k=1}^{n}a_k=\sum_{k=1}^{n}\frac{1}{k}-\frac{1}{k+1}$

$=\sum_{k=1}^{n}\frac{1}{k}-\frac{1}{k+1}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}$

$=1-\frac{1}{n+1}$

3) Hence find

$\sum_{k=1}^{\infty}a_k$

$\sum_{k=1}^{\infty}a_k=\lim_{n\to\infty}\sum_{k=1} ^{n}a_k$

$=\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)=1$

**MathoMan** · September 16th 2010, 06:18 AM

Looks fine from overe here.

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