Show that $\sum_{n=1}^{\infty} \sin \left(\frac{1}{n^2}\right)$ converges. Show that $$\sum_{n=1}^{\infty} \sin \left(\frac{1}{n^2}\right)$$ converges.
I've attempted a solution by using the following approximation of $\sin x$ for small $x$
$$
\sin x \approx x
$$
Therefore 
$$
\sin \left(\frac{1}{n^2}\right) \approx \frac{1}{n^2}
$$
So I came to the conclusion that 
$$
\sum_{n=1}^{\infty} \sin \left(\frac{1}{n^2}\right) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}
$$
Which clearly converges, however both Wolfram Alpha and my instructor claims the convergence of this series can be shown using the comparison test. However, I'm having trouble determining a sequence to compare it to.
 A: Use $\sin(x)\le x$ for $x\ge 0$.
A: You can use the fact that $\sin x<x$ when $x>0$, so that $\sin\dfrac 1 {n^2} < \dfrac 1 {n^2}$.  Then cite the "comparison test".
Your conclusion that $\displaystyle \sum_{n=1}^\infty \sin \frac 1 {n^2} = \sum_{n=1}^\infty \frac 1 {n^2}$ is too much: it should say "less than" rather than "equal to".
A: After Michael Hardy's answer, let us try to find a better bound for the sum.
Using Taylor expansion for $\sin(x)$ and replacing $x$ by $\frac 1{n^2}$,we have $$\sin( \frac 1{n^2})=\sum_{i=0}^\infty \frac{(-1)^i}{(2i+1)!}\frac 1 {n^{4i+2}}$$ $$S=\sum_{n=1}^\infty\sin( \frac 1{n^2})=\sum_{i=0}^\infty \frac{(-1)^i}{(2i+1)!}\sum_{n=1}^\infty\frac 1 {n^{4i+2}}$$ and noting $$A_i=\sum_{n=1}^\infty\frac 1 {n^{4i+2}}=\zeta ( 4 i+2)$$ with $$A_0=\frac{\pi ^2}{6}$$ $$A_1=\frac{\pi ^6}{945}$$ $$A_2=\frac{\pi ^{10}}{93555}$$ When $i$ increases $A_i\to 1$. So, as an approximation $$S\approx\frac{\pi ^2}{6}-\frac 1{3!}\frac{\pi ^6}{945}+\frac 1{5!}\frac{\pi ^{10}}{93555}+\sum_{i=3}^\infty \frac{(-1)^i}{(2i+1)!}=\frac{\pi ^2}{6}-\frac 1{3!}\frac{\pi ^6}{945}+\frac 1{5!}\frac{\pi ^{10}}{93555}+\sin (1)-\frac{101}{120}$$ This approximation gives $S\approx 1.483522829$ while the correct value should be $\approx 1.483522817$
