Calculate the integral $\int_0^{2\pi}\sum_{k=n}^{\infty}\frac{e^{i(k-m)\theta}}{k+1}d\theta$ I would like to calculate this integral:
$$\int_0^{2\pi}\sum_{k=n}^{\infty}\frac{e^{i(k-m)\theta}}{k+1}d\theta$$
where $n$ and $m$ belong to $\mathbb{N}$.
My attempt: 
Instead of considering an infinite sum, I will consider a finite one.  
Let $l>n$, we have:
\begin{alignat*}{2}
  \int_0^{2\pi}\sum_{k=n}^{l}\frac{e^{i(k-m)\theta}}{k+1}d\theta=\sum_{k=n}^{l}\frac{1}{k+1}\int_0^{2\pi}e^{i(k-m)\theta}d\theta
 \end{alignat*}
Taking the limit when $l$ tends to $\infty$, we have:
$$\lim_{l\to\infty}\int_0^{2\pi}\sum_{k=n}^{l}\frac{e^{i(k-m)\theta}}{k+1}d\theta=\left \{
   \begin{array}{l c r}
     \frac{2\pi}{m+1}  & \text{if} & m\geq n \\
     0 & \text{if} & m<n
   \end{array}
   \right. $$
In this case, do we have:
$$\lim_{l\to\infty}\int_0^{2\pi}\sum_{k=n}^{l}\frac{e^{i(k-m)\theta}}{k+1}d\theta=\int_0^{2\pi}\sum_{k=n}^{\infty}\frac{e^{i(k-m)\theta}}{k+1}d\theta\quad?$$
Many thank's in advance.
 A: Consider a branch of logarithm $$f(z):=-\ln(1-z)$$
for $z\in\mathbb{C}\setminus\mathbb{R}_{\geq 1}$ that coincides with $\displaystyle\sum_{k=1}^\infty\,\frac{z^k}{k}$ when $|z|<1$.  Define also $$f_r(z):=\sum_{k=1}^{r}\,\frac{z^k}{k}$$
for all $z\in\mathbb{C}$ and $r\in\mathbb{Z}_{\geq0}$.  It is known that $\lim\limits_{r\to\infty}\,f_r(z)=f(z)$ for all $z\in\mathbb{C}$ such that $|z|\leq1$ and $z\neq 1$.  
We now define
$$g_r(\theta):=f_r\big(\text{e}^{\text{i}\theta}\big)=\sum_{k=1}^{r}\,\frac{\text{e}^{\text{i}k\theta}}{k}$$
and
$$g(\theta):=f\big(\text{e}^{\text{i}\theta}\big)=-\ln\big(1-\text{e}^{\text{i}\theta}\big)$$
for all $\theta\in(0,2\pi)$ and $r\in\mathbb{Z}_{\geq 0}$.  Thus, $g_r\to g$ pointwise as $r\to\infty$.   For each compact subset $K$ of $(0,2\pi)$, we want to show that the family $\Big(g_r\big|_K\Big)_{r=0}^\infty$ is uniformly equicontinuous.  This proves that $g_r\big|_K\to g\big|_K$ uniformly as $r\to\infty$, and thereby, proving the OP's assertion that
$$\lim_{l\to\infty}\,\int_0^{2\pi}\,\frac{g_{l+1}(\theta)-g_n(\theta)}{\text{e}^{\text{i}(m+1)\theta}}\,\text{d}\theta=\int_0^{2\pi}\,\frac{g(\theta)-g_n(\theta)}{\text{e}^{\text{i}(m+1)\theta}}\,\text{d}\theta\,.$$
Here is the proof of the claim above.  Let $K$ be a compact subset of $(0,2\pi)$.  Fix $\epsilon>0$ and $\phi\in K$.  We want to find $\delta>0$ such that $$\big|g_r(\theta)-g_r(\phi)\big|<\epsilon\tag{*}$$ for every $r\in\mathbb{Z}_{\geq 0}$ and every $\theta\in K$ such that $|\theta-\phi|<\delta$.  However, 
$$g'_r(\theta)=\text{i}\text{e}^{\text{i}\theta}\,f_r'\left(\text{e}^{\text{i}\theta}\right)$$
so that
$$\left|g'_r(\theta)\right|=\Big|f_r'\left(\text{e}^{\text{i}\theta}\right)\Big|=\left|\frac{1-\text{e}^{\text{i}r\theta}}{1-\text{e}^{\text{i}\theta}}\right|\leq \frac{2}{\big|1-\text{e}^{\text{i}\theta}\big|}=\frac{1}{\Big|\sin\left(\frac{\theta}{2}\right)\Big|}\,.$$
Let $M$ be the supremum of $\dfrac{1}{\Big|\sin\left(\frac{\theta}{2}\right)\Big|}$ for $\theta\in K$.  Then, each $g_r$ is $M$-Lipschitz.  Therefore, for $\delta:=\dfrac{\epsilon}{M}$, we see that, whenever $\theta\in K$ satisfies $|\theta-\phi|<\delta$, (*) holds.  Since $\delta$ does not depend on $\phi$, the family $\Big(g_r\big|_K\Big)_{r=0}^\infty$ is uniformly equicontinuous.
