# Telescoping sums

#### Alexmahone

##### Active member
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 $>0$

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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#### Moo

##### New member
Hello,

****************** latex. I hate this version....

Forget it for the first one... But please post your thing at once, because you're constantly editing and it's really ticking me off to try writing something that yourself are currently writing !

As for the second one, reason as if k=2, how would you do ? Then it'd be the same except that it's a little more, but in the end they'll all cancel each other out.

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#### Alexmahone

##### Active member
As for the second one, reason as if k=2, how would you do ? Then it'd be the same except that it's a little more, but in the end they'll all cancel each other out.
I think the answer is $\frac{1}{k}(\frac{1}{1}+\ldots+\frac{1}{k})$. Can that be simplified further?

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#### Krizalid

##### Active member
Yes, that's the answer, and no further simplification follows, since the harmonic sum doesn't have much nice properties.