Does $\sum{\frac{\sin{(nx)}}{n}}$ converge uniformly for all $x$ in $[0,2\pi]$ This question arises because of a problem I was doing (Bartle 3rd edition, section 9.4 problem 3). It was like this. Given $a_n$ a decreasing sequence of positive numbers and suppose that $$\sum_{n=0}^{\infty}{a_n \sin{(nx)}}$$ Converge uniformly (It doesn't specify the domain, so I guess is for every x). Prove that $n a_n \to 0$. Clearly $\frac{1}{n}$ fits the description of $a_n$, and $n \frac{1}{n} \to 1 \neq 0$, so this would prove that there is a mistake in the problem if $\sum{\frac{\sin{(nx)}}{n}}$ converge uniformly for all $x$.
So my question is if $\sum{\frac{\sin{(nx)}}{n}}$ converge uniformly for every $x$.
(I know that the series converge uniformly for every x in $[\delta, 2\pi - \delta]$, for $0 < \delta <2\pi$ by using the Dirichlet criterion.)
 A: Hint:
Use Cauchy Criterion to prove that your infinite series isn't uniformly converges for all $x\in[0,2\pi]$.
A: The sum of the series is non-continuous (you can view this as the Fourier series for a saw-tooth function; or just check the behavior around x=0), so the convergence cannot be uniform. Each summand is obviously a continuous function, and a uniformly convergent series of continuous functions is continuous. 
A: Let $$B_n=\sum_{k=0}^nb_k,\:\:F_n(x)=\sum_{k=0}^na_kb_k,\quad a_k=k^{-1},b_k=\sin(kx),\:\:(k,n)\in\mathbb N^2$$
We have $$F_n(x)=a_0b_0+\sum_{k=1}^na_k(B_k-B_{k-1})=a_nB_n+\sum_{k=0}^{n-1}a_kB_k-\sum_{k=0}^{n-1}a_{k+1}B_k\\=a_nB_n+\sum_{k=0}^{n-1}(a_k-a_{k+1})B_k,\quad \forall k\le n-1.$$
Now, $$B_k=\mathcal Im\left(\sum_{j=0}^ke^{ij\phi}\right)=\mathcal Im\left(\frac{1-e^{i\phi(k+1)}}{1-e^{i\phi}}\right),\quad \phi\in\mathbb R,\:i\in\mathbb C.\\1-e^{i\phi}=e^{i\phi/2}\left(e^{-i\phi/2}-e^{i\phi/2}\right)=-2ie^{i\phi/2}\sin\left(\frac{\phi}{2}\right).$$
Thus, whenever $\:\phi\in\{0,2\pi\mathbb Z\},\:\:F_n\:$ becomes unbounded.
