How to see that series $|\sin(1/n^2)|$ converges or diverges? How to see that series $\sum_{n=1}^{\infty}|\sin(1/n^{2})|$ converge or diverge? That is, to see if $\sum_{n}\sin(1/n^{2})$ absolutely converges.
I know that for all $n$ in $\mathbb N$, $\sin(1/n^2)<1/n^2$, but is that fact useful here? Could someone tell me how to show that?
 A: If you're looking to prove the inequality $$\left|\sin\frac1{n^2}\right|\le\frac1{n^2},$$ it follows from:
Claim: $|\sin x|\le |x|$ for every $x$.
Proof: Suppose $x\ne0$. By the mean value theorem, there exists a number $a$ such that
$$\sin x - \sin 0 = \cos(a)(x-0),\tag1$$
since the derivative of $\sin x$ is $\cos x$. Take absolute values of both sides of (1), then use the fact that $|\cos a|\le 1$.
A: Compare with the convergent series $\sum n^{-2}$, then 
$$ \lim \frac{ \sin (n^{-2}) }{n^{-2}} =_{t = n^{-2}} \lim_{t \to 0} \frac{\sin t }{t} = 1 $$
A: Since$\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges by the p-series test, Therefore $\sum_{n=1}^{\infty}|\sin (\frac{1}{n^2})|$ converges by using the inequality mentioned by you and the comparison test.
A: For all $x$ $$\sin (x) = \sum^{\infty}_{i=0} \frac{(-1)^i}{(2i+1)!} x^{2i+1} 
$$ So $$\sin\left(\frac 1 {n^k} \right)=\sum^{\infty}_{i=0} \frac{(-1)^i}{(2i+1)!} n^{-(2i+1)k} $$ and, by comparison with p-series, each term converge as soon as $k>1$.
Moreover, you do not need to consider absolute values since $n\geq 1$.
For $k=2$, having $$\sin\left(\frac 1 {n^2} \right)=\frac{1}{n ^2}-\frac{1}{6 n^6}+\frac{1}{120
   n^{10}}+O\left(\frac{1}{n^{14}}\right)$$ $$\sum_{n=1}^\infty \sin\left(\frac 1 {n^2} \right)=\sum_{n=1}^\infty \frac{1}{n ^2}-\frac{1}{6}\sum_{n=1}^\infty \frac{1}{n ^6}+\frac{1}{120}\sum_{n=1}^\infty \frac{1}{n ^{10}}+\cdots=\frac{\pi ^2}{6}-\frac{\pi ^6}{5670}+\frac{\pi ^{10}}{11226600}$$
