convergence of a function serie to norm 2 need to show if the following function serie converge on $||.||_2$ on [1,5]:
$$\sum _1^\infty {sin^2(nx) \over n^2}  $$
I have no idea how to approach that one, would like
for some directions...
 A: We can in fact compute what the series converges to. Note that
$$\sum_{n\ge 1}\frac{\sin^2(nx)}{n^2}
= \sum_{n\ge 1}\frac{1}{2} \frac{1-\cos(2nx)}{n^2}
= \frac{\pi^2}{12} - \frac{1}{2} \sum_{n\ge 1} \frac{\cos(2nx)}{n^2}.$$
Re-write this as follows:
$$\frac{\pi^2}{12} - 2 x^2 \sum_{n\ge 1} \frac{\cos(n2x)}{(n2x)^2}.$$
Now the sum $$S(x) = \sum_{n\ge 1} \frac{\cos(nx)}{(xn)^2}$$ is harmonic and was evaluated by inverting its Mellin transform at this MSE link, the result being
$$S(x) =  \frac{\pi^2}{6x^2} -\frac{\pi}{2x} +\frac{1}{4}.$$
This gives for our sum the value
$$\frac{\pi^2}{12} - 2 x^2 
\left(\frac{\pi^2}{24x^2} -\frac{\pi}{4x} +\frac{1}{4}\right)$$
which simplifies to
$$\frac{\pi}{2} x - \frac{1}{2} x^2$$
on the interval $(0,\pi).$
A: In fact it converges uniformly.  Let $f_n(x)=a_nsin^2(nx)$ with $a_n=1/n^2$. Then $\|f_n\|_{\infty}=max\|f_n(x)|=|a_n|$.  
We have that $\sum_{n=1}^\infty |a_n|$ converges (exactly to $π^2/6$) and by Weierstrass Criterion of series convergance wwe have that $\sum_{n=1}^\infty \|f_n\|_{\infty}$ converges and thus $\sum_{n=1}^\infty f_n$ converges uniformly on $\Bbb R$.
