I already showed that $\displaystyle g(x)=\sum_{n=1}^{\infty} \frac{\sin(nx)}{n}$ converges uniformly on $[\delta,2\pi-\delta ]$ for $\delta>0$
Now I have to calculate the limit of $\displaystyle\lim_{x\to 1^-}\sum_{n=1}^{\infty}\frac{x^n*exp(int)}{n}$ for $t\in (0,2\pi)$
For $\displaystyle f(t)=\sum_{n=1}^{\infty}\frac{exp(int)}{n}$ I showed convergence, so I guess $\displaystyle F(x)=\sum_{n=1}^{\infty}\frac{exp(int)}{n}x^n$ converges for $x=1$. Is this enough to say that that $f(t)$ is the limit of $F(x)$ as $x\to1^-$?
I would continue with Abel's theorem that $F(x)$ is a continuous function on $[0,1]$.
Here is where I don't make any progress. I should calculate the result of $F(x)$ and then conclude to the result of $g(x)$. I guess that the imaginary part will be $g(x)$.
Are my thoughts right? How can I calculate the result? Any hints?
$f(t)$ does not converge uniformly on $[0, 2\pi]$! At $t=0$ and $x\nearrow 1$ you have $F(x) = \infty$, thus $F$ cannot be continuous on $[0,1]$