How to figure $\displaystyle\sum_{n=1}^{\infty}\sin\frac{1}{n^2}$? $\displaystyle\sum_{n=1}^{\infty}\sin\frac{1}{n^2}=?$
Does it have a precise expression?
 A: The behavior of Sinusoidal functions like Sine and Cosine is not always well defined especially when $x\to \infty$, because of their oscillatory behavior. 
If you are in search of a close numerical form, yes it does exist but it has been obtained through numerical methods and not analytical ones (like for example other well known series).
Anyway by comparison test, the series does converge (absolutely), and it's numerical convergence is
$$1.483522817221346$$
A: Use the limit comparison test:
$$\frac{\left|\sin\frac1{n^2}\right|}{\frac1{n^2}}\xrightarrow[n\to\infty]{}1$$
and thus our series converges absolutely . By the way, you can drop the absolute value above. Can you see why?
A: Since $\sin x \approx x$ when $x$ is small and $1/n^2 \to 0$ you might expect the series to be reasonably similar to 
$$\sum_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} = \approx 1.645$$. 
You can at least use this series as an upper bound of your series to establish convergence in a pretty straightforward manner. Inspired by this I consider the Maclaurin expansion
$$\sin \left ( \frac{1}{n^2} \right ) = \sum_{k = 0}^\infty \frac{(-1)^k}{(2k + 1)!}\frac{1}{n^{2(2k + 1)}}$$ and 
$$\sum_{n = 1}^\infty \sin \left ( \frac{1}{n^2} \right ) = \sum_{n = 1}^\infty  \sum_{k = 0}^\infty \frac{(-1)^k}{(2k + 1)!}\frac{1}{n^{2(2k + 1)}} = \sum_{k = 0}^\infty \frac{(-1)^k}{(2k + 1)!} \sum_{n = 1}^\infty \frac{1}{n^{2(2k + 1)}} $$
which is suggestive of 
$$\sum_{n = 1}^\infty \sin \left ( \frac{1}{n^2} \right ) =  \sum_{k = 0}^\infty \frac{(-1)^k}{(2k + 1)!} \zeta(2(2k + 1)) $$
which I think is cute because there is a closed formula for the zeta function of the even numbers 
$$\zeta(2n ) = \frac{(-1)^{n + 1}B_{2n}(2\pi)^{2n}}{2(2n)!}$$
where $B_{k}$ is a bernoullinumber. I'm typing this pretty quickly so I'm not really checking convergences and I am a bit worried that the wolfram summation doesnt agree with its own for sin very well but that could just be rounding errors. 
If anything it converges extremely quickly even if it isn't 'closed'
EDIT: If we 'simplify' we get the sum
$$\sum_{n = 0}^\infty \frac{(-1)^n B_{2(2n + 1)}(2\pi)^{2(2n  +1)}}{2(2n + 1)![2(2n + 1)]!}$$
