Limit of sum of $\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^2}$ find the lim
$$\lim_{x\to 0}\; \sum_{n=1}^{\infty}\frac{\sin(nx)}{n^2}$$  
Is it correct to do this:
$$\lim_{x\to 0} x\frac{\sin x}{x}+x\frac{\sin2x}{4x}+x\frac{\sin3x}{9x} + \cdots$$
this becomes
$$\lim_{x\to 0} x+\frac{x}{2}+\frac{x}{3}+\cdots $$
or
$$\lim_{x\to 0} x\left(1+\frac{1}{2}+\frac{1}{3} + \cdots \right) $$
 now what does this result to?
if this way is wrong, then can we first find sum of $\displaystyle\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^2}$  , how to proceed to find this sum?
 A: We have
$$\left|\frac{\sin(nx)}{n^2}\right|\le\frac1{n^2}$$
and the series $\sum_{n\ge1}\frac 1{n^2}$ is convergent so by the Weierstrass M-test  the given series is uniformly convergent on $\mathbb R$ hence
$$\lim_{x\to0}\sum_{n=1}^\infty \frac{\sin(nx)}{n^2}=\sum_{n=1}^\infty \lim_{x\to0}\frac{\sin(nx)}{n^2}=0$$
A: You can do it "from scratch" as follows. 
Of course, this will only mimic the proof of a general "interchange of limit with a sum" theorem (namely the dominated convergence theorem for series), so that Sami's answer is arguably "better" than this one.
Set $f(x):=\sum_1^\infty \frac{\sin(nx)}{n^2}\cdot$ Let us show that $f(x)\to 0$ as $x\to 0$. 
Start with $\varepsilon >0$. Since the series $\sum\frac{1}{n^2}$ is convergent, you can find $n_0$ such that $\sum_{n>n_0}\frac1{n^2}\leq \varepsilon$. Since $\vert\sin(nx)\vert\leq 1$, you then have (by the triangle inequality)
\begin{eqnarray}\forall x\in\mathbb R\;:\; \vert f(x)\vert&\leq& \left\vert\sum_{n=1}^{n_0} \frac{\sin (nx)}{n^2}\right\vert+\sum_{n>n_0}\frac1{n^2}\\&\leq& \left\vert\sum_{n=1}^{n_0} \frac{\sin (nx)}{n^2}\right\vert+\varepsilon\, .
\end{eqnarray}
Now, the sum in the right-side (call it $S_{n_0}(x)$) is a finite sum of terms all tending to $0$ as $x\to 0$. Hence $S_{n_0}(x)\to 0$ as $x\to 0$, so you can find $\delta>0$ such that $\vert x\vert<\delta\implies \vert S_{n_0}(x)\vert\leq\varepsilon$. Therefore, you get the implication
$$\vert x\vert<\delta\implies \vert f(x)\vert\leq \varepsilon+\varepsilon=2\varepsilon\, .$$
This shows that $f(x)\to 0$ as $x\to 0$, as required.
