Is the series $\sum_{n=0}^\infty(\frac{n}{2n^2-1})^2$ convergent?
My attempt:
The series is convergent iff $\sum_{n=0}^\infty 2^ka_{2^k}$ convergent. Then we need to show that $\sum_{n=0}^\infty 2^k (\frac{2^k}{2\times2^k-1})^2$ converges. I want to find a series that bigger than the above series, but don't know what to do. Can someone help me with that? Or does root test works in this case?
Yes, of course because for $n\rightarrow+\infty$ $$\frac{\frac{n^2}{(2n^2-1)^2}}{\frac{1}{n^2}}\rightarrow\frac{1}{4}$$ and $\sum\limits_{n=1}^{+\infty}\frac{1}{n^2}$ converges (to $\frac{\pi^2}{6}$).
Also, we have $$\sum_{n=2}\frac{n^2}{(2n^2-1)^2}<\sum_{n=2}^{+\infty}\frac{1}{n(n+1)}=\sum_{n=2}^{+\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1}{2}$$