# Sum of the series

• March 22nd 2013, 07:58 AM
Shakarri
Sum of the series
I am struggling to see a way of finding the infinite sum of the series

$2sin(\frac{1}{k}-\frac{1}{k+1})cos(\frac{1}{k}+\frac{1}{k+1})$

$sin(\frac{1}{k}-\frac{1}{k+1})$ converges to 0 and $cos(\frac{1}{k}+\frac{1}{k+1})$ converges to 1.

Expanding and simplifying the trig functions just yields
$2(sin(\frac{1}{k})cos(\frac{1}{k})cos(\frac{2}{k+1 })-sin(\frac{1}{k+1})cos(\frac{1}{k+1})cos(\frac{2}{k }))$

Which is not of any use

We didn't cover many methods for finding the sum of series in class, just ways of showing that a series converges :(
Could someone point me in the right direction?
• March 22nd 2013, 05:25 PM
Nehushtan
Re: Sum of the series
$2\sin\left(\frac1k - \frac1{k+1}\right)\cos\left(\frac1k + \frac1{k+1}\right)\,=\,\sin\frac2k-\sin\frac2{k+1}$

$\sum_{k=1}^n\left[\sin\frac2k-\sin\frac2{k+1}\right]\,=\,\sin2-\sin\frac2{n+1}\to\sin2\ \mbox{as}\ n\to\infty$
• March 23rd 2013, 09:07 AM
Shakarri
Re: Sum of the series
At first I didnt see what you did but now I see its a telescoping series! thank you