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Telescoping sums

Alexmahone

Active member
Jan 26, 2012
268
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?
 
Last edited:

Moo

New member
Feb 12, 2012
26
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.
 
Last edited by a moderator:

Alexmahone

Active member
Jan 26, 2012
268
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
Feb 9, 2012
118
Yes, that's the answer, and no further simplification follows, since the harmonic sum doesn't have much nice properties.