# 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 )

• If Dirichlet's test were allowed, I would say rewrite $\sin^2 n = \frac{1}{2}(1-\cos(2n))$ and then you get the sum of a divergent series and a convergent series. May 2, 2019 at 22:58
• @DonThousand but this required complex analysis which is above the question level May 2, 2019 at 23:11
• @DonThousand I don't see the relevance to this question. $\frac{\sin^2 x}{x}$ isn't a decreasing function so the integral test won't be applicable - and for example, $\int_1^n \frac{\sin^2(\pi x)}{x} dx$ is also asymptotic to $\frac{1}{2}\log(n)$. May 2, 2019 at 23:19

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 $$b_n.
Consider \begin{align} \sum_{n=1}^{\infty}\frac{\sin^2(n)}{n}& >\sum_{n=1}^{\infty}\frac{\sin^2(a_n)}{a_n}&&\text{just summing over fewer positive terms}\\ &>\frac12\sum_{n=1}^{\infty}\frac{1}{a_n}&&\text{property of a_n}\\ &\geq\frac12\sum_{n=1}^{\infty}\frac{1}{b_n}&&\text{just summing over fewer positive terms}\\ &>\frac1{2\pi}\sum_{n=1}^{\infty}\frac{1}{n}&&\text{property of b_n} \end{align}
• I think, there was no need to define the $a_n$. You can just choose $b_n$ to be an integer in the subinterval in the middle. May 3, 2019 at 16:41
• @JulianMejia If you are going for a most streamlined argument, yes. It was more natural (to me) to first consider the existence of the $a_n$, and then get a hold on their somewhat unpredictable frequency in a second stage. Sometimes I prefer a more natural argument over a streamlined one. May 3, 2019 at 18:48
Sorry this would be only a comment if it weren't for my low reputation since it is at best a proof sketch, but I believe we may have (in my class) looked at all the natural numbers that are better and better approximations of $$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}$$.