How to prove that $\sum\limits_{n=1}\frac{\sin^2n}n$ is divergent How to prove that $\sum\limits_{n=1}\frac{\sin^2n}n$ does not converge without using the series expansion and with the following tests only or a combination of them (comparison or limit comparison test )
 A: Sorry this would be only a comment if it weren't for my low reputation since it is at best a proof sketch, but I believe we may have (in my class) looked at all the natural numbers that are better and better approximations of $2\pi k+\frac{\pi}{2}$, which makes the top strictly greater than the first one in that sequence, and found that subsequence to be divergent since it was a not-too-horrible-to-analyze divergent subsequence of $\sum_{i=1}^{+\infty} \frac{k}{n}$.
A: There is a sequence of positive integers $\{a_n\}=\{1,2,4,5,8,\ldots\}$ where for each $a_n$ in this sequence, $\sin^2a_n$ is greater than $\frac12$, and this is the maximal such sequence.
How frequent are these $a_n$? In every interval $[(k-1)\pi,k\pi]$, there is a subinterval in the middle of length $\pi/2$ where $\sin^2(x)>\frac12$. Since $\pi/2>1$, there is always an integer in this subinterval. So in the intervals $[0,\pi],[\pi,2\pi],[2\pi,3\pi],\ldots$ you can always find (at least) one $a_n$. Let $b_n$ be some integer in the intersection of $[(n-1)\pi,n\pi]$ with $\{a_n\}$. Note $b_n<n\pi$.
Consider $$
\begin{align}
\sum_{n=1}^{\infty}\frac{\sin^2(n)}{n}&
>\sum_{n=1}^{\infty}\frac{\sin^2(a_n)}{a_n}&&\text{just summing over fewer positive terms}\\
&>\frac12\sum_{n=1}^{\infty}\frac{1}{a_n}&&\text{property of $a_n$}\\
&\geq\frac12\sum_{n=1}^{\infty}\frac{1}{b_n}&&\text{just summing over fewer positive terms}\\
&>\frac1{2\pi}\sum_{n=1}^{\infty}\frac{1}{n}&&\text{property of $b_n$}
\end{align}$$
