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#### Alexmahone

##### Active member

- Jan 26, 2012

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Evaluate using telescoping sums:

(a) $\sum_1^\infty\frac{(-1)^{n-1}}{n(n+2)}$

(b) $\sum_1^\infty\frac{1}{n(n+k)}$, $k$ integer $>0$

(a)$\frac{1}{n(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)$

Adding the terms for $n$ even, we get

$-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\ldots\right)=-\frac{1}{4}$

Adding the terms for $n$ odd, we get

$\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\ldots\right)=\frac{1}{2}$

So, the total is $-\frac{1}{4}+\frac{1}{2}=\frac{1}{4}$

(b) $\sum_1^\infty\frac{1}{n(n+k)}=\sum_1^\infty\frac{1}{k}\left(\frac{1}{n}-\frac{1}{n+k}\right)$

$=\frac{1}{k}\left(\frac{1}{1}-\frac{1}{1+k}+\frac{1}{2}-\frac{1}{2+k}+\frac{1}{3}-\frac{1}{3+k}+\ldots\right)$

How do I proceed?

(a) $\sum_1^\infty\frac{(-1)^{n-1}}{n(n+2)}$

(b) $\sum_1^\infty\frac{1}{n(n+k)}$, $k$ integer $>0$

**My attempt:**(a)$\frac{1}{n(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)$

Adding the terms for $n$ even, we get

$-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\ldots\right)=-\frac{1}{4}$

Adding the terms for $n$ odd, we get

$\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\ldots\right)=\frac{1}{2}$

So, the total is $-\frac{1}{4}+\frac{1}{2}=\frac{1}{4}$

(b) $\sum_1^\infty\frac{1}{n(n+k)}=\sum_1^\infty\frac{1}{k}\left(\frac{1}{n}-\frac{1}{n+k}\right)$

$=\frac{1}{k}\left(\frac{1}{1}-\frac{1}{1+k}+\frac{1}{2}-\frac{1}{2+k}+\frac{1}{3}-\frac{1}{3+k}+\ldots\right)$

How do I proceed?

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