How to calculate the limit of the following sum as it approaches a divergence I am interested in the behaviour of summations like the following:
$$
\lim_{\alpha\to 0}\sum_{n=1}^{\infty}\frac{e^{in\alpha}}{\sqrt{\beta+n^{2}}}
$$
where $\beta$ is a constant. Just putting $\alpha=0$ doesn't help because the sum doesn't converge (it has a logarithmic singularity). I know that the case $\beta=0$ is easily calculated because the summation can be done exactly:
$$
\sum_{n=1}^{\infty}\frac{e^{in\alpha}}{n}=-ln(1-e^{i\alpha})\sim\frac{\pi i}{2}-ln(\alpha)+O(\alpha)
$$
but I don't know how to extend this to the case when $\beta$ is non-zero.
Thanks in advance for any help.
 A: What about Euler Mac-Laurin?
Let us assume that $\alpha,\beta >0$ and denote the sum in question by $S_{\alpha,\beta}$.
For general  $\alpha$ we would get (we deal with the sum starting at $n=0$)
$$ S_{\alpha,\beta}\sim  I_{\alpha,\beta}  + \frac{1}{2\sqrt{\beta}}+R_{\alpha,\beta}^p =\int_0^{\infty}dx \frac{e^{i\alpha x}}{(x^2+\beta)^{1/2}}+\frac{1}{2\sqrt{\beta}}+\sum_{k=1}^{p}\frac{d^{2k-1}}{dx^{2k-1}}\frac{e^{i\alpha x}}{(x^2+\beta)^{1/2}}|_{x=0}$$
The integral is well known as a linear combination of Struve and modified Besselfunctions
$$
I_{\alpha,\beta} = K_0(\alpha\sqrt{\beta})-\frac{i\pi}2(I_0(\alpha\sqrt{\beta})-L_0(\alpha\sqrt{\beta})
$$
Using asymptotics for this well known functions we obtain in the limit $\alpha \rightarrow 0_+$
$$
I_{\alpha,\beta}\sim -\log(\alpha\beta)-\gamma+\frac{i \pi}{2}+o(1)
$$
also, it is not difficult to show that, as $\alpha \rightarrow 0_+$
$$
R_{\alpha,\beta}^p \sim o(1)
$$
So finally
$$
S_{\alpha,\beta} \sim -\log(\alpha)-\frac{\log(\beta)}2+\frac{1}{2\sqrt{\beta}}-\gamma+\log(2)+\frac{i \pi}{2}+o(1)
$$
Which matches pretty well numerically
Not sure how to include your limit $\beta \rightarrow 0$ consistently. We can just subract the $0$ term from our initial sum but this doesn't seem enough. Note also, that you can get arbitrary good results from corrections out of the remainder term ($2i a$ should be the next order).
A: I will assume that $\alpha>0$ and $\beta \geq 0$. Let us write
$$
\sum\limits_{n = 1}^\infty  {\frac{{e^{in\alpha } }}{{\sqrt {\beta  + n^2 } }}}  = \sum\limits_{n = 1}^\infty  {\frac{{e^{in\alpha } }}{n}}  + \sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{\sqrt {\beta  + n^2 } }} - \frac{1}{n}} \right)e^{in\alpha } } .
$$
As you noted,
$$
\sum\limits_{n = 1}^\infty  {\frac{{e^{in\alpha } }}{n}}  =  - \log \alpha  + \frac{\pi }{2}i + \mathcal{O}(\alpha )
$$
as $\alpha\to 0+$. Now,
\begin{align*}
\sum\limits_{n = 1}^\infty\! {\left( {\frac{1}{{\sqrt {\beta  + n^2 } }} - \frac{1}{n}} \right)e^{in\alpha } }  =\; & \sum\limits_{n = 1}^\infty \! {\left( {\frac{1}{{\sqrt {\beta  + n^2 } }} - \frac{1}{n}} \right)} \\ & + \sum\limits_{n = 1}^\infty  {(e^{in\alpha }  - 1)\!\left( {\frac{1}{{\sqrt {\beta  + n^2 } }} - \frac{1}{n}} \right)} .
\end{align*}
It is easy to see that
$$
0 < \frac{1}{n} - \frac{1}{{\sqrt {\beta  + n^2 } }} = \frac{1}{2}\int_0^\beta  {\frac{{dx}}{{(x + n^2 )^{3/2} }}} 
\le \frac{\beta }{{2n^3 }},
$$
whence
\begin{align*}
\left| {\sum\limits_{n = 1}^\infty  {(e^{in\alpha }  - 1)\left( {\frac{1}{{\sqrt {\beta  + n^2 } }} - \frac{1}{n}} \right)} } \right| & \le \sum\limits_{n = 1}^\infty  {\left| {\frac{1}{{\sqrt {\beta  + n^2 } }} - \frac{1}{n}} \right|\left| {e^{in\alpha }  - 1} \right|}\\ &  \le \beta \sum\limits_{n = 1}^\infty  {\frac{1}{{n^3 }}\left| {\sin \left( {\frac{{n\alpha }}{2}} \right)} \right|}  \le \beta \alpha \frac{{\pi ^2 }}{{12}}
\end{align*}
and
$$\left|\sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{\sqrt {\beta  + n^2 } }} - \frac{1}{n}} \right)}\right|<+\infty.$$
In summary,
$$
\sum\limits_{n = 1}^\infty  {\frac{{e^{in\alpha } }}{{\sqrt {\beta  + n^2 } }}}  =  - \log \alpha + C_\beta   + \frac{\pi }{2}i  + \mathcal{O}(\alpha ) + \mathcal{O}(\alpha \beta )
$$
as $\alpha \to 0+$, uniformly in $\beta\geq 0$ with
$$
C_\beta   = \sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{\sqrt {\beta  + n^2 } }} - \frac{1}{n}} \right)} .
$$
A simplified result which is useful for larger values of $\beta$ may be obtained as follows. It can be shown using the Euler–Maclaurin formula that
$$
C_\beta   =  - \log \sqrt{\beta}  - \gamma  + \log 2 + \mathcal{O}\!\left( {\frac{1}{{\sqrt {\beta+1}  }}} \right).
$$
Therefore,
$$
\sum\limits_{n = 1}^\infty  {\frac{{e^{in\alpha } }}{{\sqrt {\beta  + n^2 } }}}  =  - \log (\alpha\sqrt{\beta})  - \gamma  + \log 2 + \frac{\pi }{2}i  + \mathcal{O}(\alpha ) + \mathcal{O}(\alpha \beta ) + \mathcal{O}\!\left( {\frac{1}{{\sqrt {\beta+1}  }}} \right) 
$$
as $\alpha \to 0+$, uniformly in $\beta\geq 0$.
