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.