Uniform convergence of complex series with $|z|=1$ but $z\neq 1$. Prove that $\sum\limits_{k=0}^\infty\frac{z^k}{k+1}$ converges where $|z|=1$ but $z\neq 1$. 
This gives an example of a power series with radius of convergence 1 that converges at every point of the unit circle except at $z=1$. The series may be written as
$$\sum\limits_{k=1}^\infty\frac{e^{ikt}}{k}.$$
Prove that for each $a\in(0,\pi)$, its convergence is uniform for $a<t<2\pi-a$.
For the first part, a previous exercise show that $\sum\limits_{k=0}^\infty a_kb_k$ converges if:
(i) The partial sums of the $a_k$'s are bounded.
(ii) $\lim\limits_{k\to\infty}b_k=0$
(iii) $\sum\limits_{k=0}^\infty|b_k-b_{k+1}|<\infty$
I think (ii) is obvious.
For (i) Since $S_N=\sum\limits_{k=0}^N z^k$ and $z\neq 1$, we have that $|1-z|=M>0$ and
$|S_N|=\bigg|\frac{1-z^{k+1}}{1-z}\bigg|=\frac{\big|1-z^{k+1}\big|}{M}\leq\frac{|1|+|z|^{k+1}}{M}=\frac{2}{M}.$
For (iii) $\sum\limits_{k=0}^\infty\bigg|\frac{1}{k+1}-\frac{1}{k+2}\bigg|=\sum\limits_{k=0}^\infty\bigg|\frac{1}{(k+1)(k+2)}\bigg|\leq\sum\limits_{k=0}^\infty\frac{1}{k^2}<\infty$ since the last series is a $p$-series.
Since all three conditions hold, for $|z|=1$ and $z\neq 1$, $\sum\limits_{k=0}^\infty\frac{z^k}{k+1}$ converges.
I'm stuck on the second part. I'm not sure how to show the uniform convergence. Since I don't know what it converges to, I thought I might use the Weierstrass M Test, but
$\big|\frac{e^{ikt}}{k}\big|\leq\frac{1}{k}$ and $\sum\limits_{k=1}^\infty\frac{1}{k}$ diverges. 
 A: If you multiply the original series with $z-1$, you get a series that converges uniformly on the entire closed unit disk. This shows that the original series converges uniformly on the closed unit disk minus a neighbourhood of 1.
A: Define
$$ S_N(t)=\sum_{k=1}^N\frac{e^{ikt}}{k}. $$
Then
\begin{eqnarray}
S_N'(t)&=&i\sum_{k=1}^N e^{ikt},\\
&=&i\frac{e^{it}(1-e^{iNt})}{1-e^{it}}
\end{eqnarray}
and hence for $a\le t\le 2\pi-a$, we have
\begin{eqnarray}
S_N(t)-S_N(a)&=&i\int_{a}^t\frac{e^{it}(1-e^{iNx})}{1-e^{ix}}dx\\
&=&i\int_{a}^t\frac{e^{ix}}{1-e^{ix}}dx-i\int_{a}^t\frac{e^{i(N+1)x}}{1-e^{ix}}dx\\
&=&\ln(1-e^{it})-\ln(1-e^{ia})-\frac{1}{N+1}\int_{a}^t\frac{1}{1-e^{ix}}de^{i(N+1)x}\\
&=&\ln(1-e^{it})-\ln(1-e^{ia})-\frac{1}{N+1}\left(\frac{e^{i(N+1)x}}{1-e^{ix}}\bigg|_a^t+\int_{a}^t\frac{e^{i(N+2)x}}{(1-e^{ix})^2}dx\right)\\
\end{eqnarray}
Note that $a\le x\le t\le 2\pi-a$ and hence 
$$\frac a2\le \frac x2\le \frac t2\le \pi-\frac a2.$$ 
Thus $|1-e^{ix}|=2|\sin \frac x2|\ge 2\sin\frac a2>0$ and hence
\begin{eqnarray}
&&\bigg|\frac{e^{i(N+1)x}}{1-e^{ix}}\bigg|_a^t+\int_{a}^t\frac{e^{i(N+2)x}}{(1-e^{ix})^2}dx\bigg|\\
&\le& \bigg|\frac{e^{i(N+1)x}}{1-e^{ix}}\bigg|_a^t\bigg|+\bigg|\int_{a}^t\frac{e^{i(N+2)x}}{(1-e^{ix})^2}dx\bigg|\\
&\le&\frac{1}{\sin\frac a2}+\frac{t-a}{4\sin^2\frac a2}\\
&\le&\frac{1}{2\sin\frac a2}+\frac{\pi-a}{2\sin^2\frac a2}\\
&<&\infty.
\end{eqnarray}
Thence
$$ S_N(t)-S_N(a)\to \ln(1-e^{it})-\ln(1-e^{ia}) \text{ uniformly as }N\to\infty. $$ 
