convergence $\sum_{n=1}^{\infty}\frac{1}{n}\sin\left(\frac{\pi}{n^{2}}\right)$ I'm checking convergence of the series
$\sum_{n=1}^{\infty}\frac{1}{n}\sin\left(\frac{\pi}{n^{2}}\right)$
using the integral test.
I calculated the integral $\int_{1}^{+\infty}\frac{1}{x}\sin\left(\frac{\pi}{x^{2}}\right)dx$ using substitution $u=\frac{\pi}{x^{2}}$,
I got: $\frac{1}{2}\int_{0}^{\pi}\frac{\sin u}{u}du$
but I don't know what to do next
thanks for any help, and sorry if I have English mistakes.
 A: It can be proven that $|\sin\alpha|\le|\alpha|$ for any real $\alpha$, so we have
$$
\sum_{n\ge1}\frac1n\left|\sin\left(\pi\over n^2\right)\right|\le\sum_{n\ge2}{\pi\over n^3}=\pi\zeta(3)-\pi
$$
Thus the series converges absolutely.
In fact, you can plug the aforementioned inequality into the integral to get a similar result.
A: Using the following fact:
$$-1 \leq \sin(x) \leq 1 \Rightarrow |\sin(x)| \leq 1$$
and it becomes evident that:
$$\sum_{n=1}^{\infty}\frac{1}{n}\sin\left(\frac{\pi}{n^{2}}\right) \leq \sum_{n=1}^{\infty} \frac{1}{n}\left |\sin \left(\frac{\pi}{n^{2}}\right)\right| \le \sum_{n = 1}^{\infty} \frac{1}{n}$$
But the p series, since $p \leq 1$ the test isn't conclusive, thus we need something else. It's possible to prove that:
$$|\sin(\pi/n) - \sin(\pi/(n+1))| \leq C /n^2$$ for some constant $C > 0$, then, it's clear that, it happens the following:
$$\Big|\sum_{n=k}^l\sin(\pi/2n)-\sin(\pi/(2n+1))\Big|\leq \frac{C}{4}\sum_{n=k}^ln^{-2}\xrightarrow{k,l\rightarrow \infty}0,$$
so the sequence is Cauchy. Well, since
$$\sum_{n=1}^{\infty}\frac{1}{n}\sin\left(\frac{\pi}{n^{2}}\right) \leq \sum_{n=1}^{\infty} \frac{1}{n}\left |\sin \left(\frac{\pi}{n^{2}}\right)\right| \leq \sum_{n=1}^{\infty} \frac{1}{n}\left |\sin \left(\frac{\pi}{n^{}}\right)\right|   $$
And since, for som $k, l \to \infty$, we have that the difference between the terms will be lesser than $\frac{C}{n^2} * \frac{1}{n(n-1)}$, but by the argument that $$\sum_{n = 2}^{\infty} \frac{C}{n^3(n-1)} \leq \sum_{n = 2}^{\infty} \frac{C}{n^2} $$
converges (because of the $p$ series convergence the $1/n^2$ series), then we can say that the original series will converge.
