Evaluate Telescoping Sums:a/b for Integer k>0

[alexmahone]

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?

 

[Moo1]

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.

 

[alexmahone]

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?

 

[Krizalid1]

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