Checking convergence of $\sum\frac{\sin nx}{n}$ Consider the sequence $$f_n(x)=\sum_{k=1}^{n}\frac{\sin kx}{k}\quad x\in \mathbb{R}$$ now we have to check convergence of $\{f_n\}$. 
Now, I used Dirichlet's criterion to show that $f_n$ converges pointwise. But uniform convergence or absolute convergence of this sequence, I cannot conclude. Can someone tell me how to look for it? thanks.
 A: More is true:
If (an ) is non-negative and decreasing, then the series $\sum\limits_{n = 1}^\infty  {{a_n}\sin (nx)} $
   is iniformly convergent  in R if and only if  n an tends to 0.
You can show the converse by just checking on the interval [0,π].
On this interval it converges to a continuous function g(x), by assuming uniform convergent you can integrate g term by term to get
$\sum\limits_{n = 1}^\infty  {{a_n}} \frac{1}{n}(1 - \cos (nx))$.  Note that g is 0 at 0 and is continuous at 0. Then for small x say x = π / k, |g(x)| < ε for |x| < δ. So 
$\int_0^\theta  {g(x)dx \le \theta \varepsilon } $ for 0 < θ <δ. Now choose π / k < δ . You can then show that${a_k} < \sum\limits_{n = 1}^k {\frac{{{a_n}}}{n}(1 - \cos (n\pi /k) < } \sum\limits_{n = 1}^\infty  {\frac{{{a_n}}}{n}(1 - \cos (n\pi /k) = \varepsilon \pi /k} $. From here you can conclude that k ak tends to 0.
Since k 1/k = 1 does not tend to 0 , the series is not uniformly convergent in [0, π] and so not uniformly convergent in R.
See Exercise 21 Question 26 in
https://037598a680dc5e00a4d1feafd699642badaa7a8c.googledrive.com/host/0B4HffVs7117IbmZ2OTdKSVBZLVk/MA3110/Chapter%209%20Uniform%20Convergence%20and%20Integration.pdf 
