Convergence of series of functions: $f_n(x)=u_n\sin(nx)$ 
Let $f_n(x)=u_n\sin(nx)$  where $\displaystyle\sum f_n$  converges pointwise, and $ \displaystyle x \mapsto \sum_{n=0}^{+\infty} f_n(x)$ is continuous.
Prove that $ u_n\rightarrow 0$ when n tends to $+\infty$

My 'attempt':
We know that  $S_n=\displaystyle\sum_{k=1}^{n}u_k\sin(kx)$ converges to a real $a$, and 
$$
S_{n+1}-S_n=u_{n+1}sin((n+1)x) 
$$
converges to $0$.
By induction we can prove that for all $x=(\frac{\pi}2)^k$ then $u_n$ tends to $0$.
I really don't know how can I continue.
Thank you in advance,
 A: It's much simpler than my comments above indicate.
All you need is that the sum converges pointwise  on a non-degenerate interval $[a,b]$ (in fact, all you need is that the sum converges on a set of positive measure; see my last comment above). Assume this is the case.
Suppose $\lim\limits_{n\rightarrow\infty} u_n\ne0$. Choose $\delta>0$ and a subsequence $(u_{n_k})$ such that $|u_{n_k}|>\delta$ for all $k\in\Bbb N$.
Note we must have 
$$\lim_{k\rightarrow\infty}u_k\sin(k x)= 0\ \text{ for all }\ x\in[a,b]\tag{1}$$
Given any closed interval $I\subset [a,b]$,
 we can find $n_k$ with $k$ as large as desired and $n_k|I|>2\pi$. With $n_k$ selected as such, we can find a closed subinterval of $J$ of $I$ so that $ \sin(n_k x)>1/2$ for all $x\in J$.
Using this, we  inductively find a sequence of closed intervals $[a,b]\supset I_1\supset I_2\supset\cdots$ and positive integers $n_1<n_2<\cdots$ such that
$$
|u_{n_k}|\sin(n_kx)\ge \delta/2\tag{2}
$$
for $x\in I_k$, $k\in \Bbb N$.
The set $I=\bigcap\limits_{k=1}^\infty I_k$ is non-empty.  Pick $x\in I$. Then $x\in[a,b]$  and satisfies $(2)$ for all $k\in\Bbb N$.  But this contradicts $(1)$.
It follows that $\lim\limits_{n\rightarrow\infty} u_n=0$.

(I don't see how to obtain a simpler proof that takes advantage of your stronger hypothesis.)
