# Thread: comparison test sum exists

1. ## comparison test sum exists

HI,

I know
$\displaystyle \sum _{k=1}^{\infty }{sin (k)}{k}^{-2}$ exists. what about

$\displaystyle \sum _{k=1}^{\infty }{cos (k)}{k}^{-2}$ does this exist?

Also how can i prove the sum of above two exists?

Thanks

2. How did you prove that the first one exists? can you use something similar to prove the second exists?

The sum of two series exists if both series exists and the proof is really simple.

3. Sin (k) and Cos (k) are bouned by 1.
the sum of K^{-2} converges absolutely.

4. Originally Posted by charikaar
HI,

I know
$\displaystyle \sum _{k=1}^{\infty }{sin (k)}{k}^{-2}$ exists. what about

$\displaystyle \sum _{k=1}^{\infty }{cos (k)}{k}^{-2}$ does this exist?

Also how can i prove the sum of above two exists?

Thanks
I mean, honestly the best way to do it would be as Shanks said and as Jose27 was undoubtedly hinting at. But for a more convoluted way using only the fact that $\displaystyle \sum_{n=1}^{\infty}\frac{\sin(n)}{n^2}$ converges is to note that $\displaystyle \cos(2x)=1-2\sin^2(x)$ so that $\displaystyle \sum_{n=1}^{\infty}\frac{\cos(n)}{n^2}=\sum_{n=1}^ {\infty}\frac{1}{n^2}-2\sum_{n=1}^{\infty}\frac{\sin^2\left(\frac{n}{2}\ right)}{n^2}$. The result follows using an obvious comparison.