[Math] Series convergence of $\sin(\frac \pi{n^2})$

Question

I would like to know how, using the limit comparison test, to show that the series $$\sum_{n=1}^{\infty} \sin \left(\frac{\pi}{n^2}\right)$$ converges.

(The ratio test in inconclusive. And the integral test shows convergence; but I know this only because I used a calculator because the antiderivative is not an elementary function.)

So if $$a_n= \sin \left(\frac{\pi}{n^2}\right)$$, then I need to find a convergent series $$\sum_{n=1}^{\infty} b_n$$ such that $$0 <= a_n <= b_n$$

Answer 1

Note that $0\le\sin(x)\le x$ for $0\le x\le\pi$. Thus, $$ \sum_{n=1}^\infty\sin\left(\frac\pi{n^2}\right)\le \pi\sum_{n=1}^\infty\frac1{n^2} $$ so the series converges by comparison with the series on the right (which converges by the $p$-test).