Asymptotics for simple series For $x > 0$, let
$$
f(x)=\sum_{n=1}^{\infty} \frac{1}{n^2+x}
$$
Can anyone find a simple equivalent of $f(x)$ when $x\to+\infty$ ?
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$x > 0$

\begin{align}
\sum_{n = 1}^{\infty}{1 \over n^{2} + x}
&=
-\,{1 \over x}
+
\sum_{n = 0}^{\infty}{1 \over \pars{n + \ic\,x^{1/2}}\pars{n - \ic\,x^{1/2}}}
=
-\,{1 \over x}
+
{\Psi\pars{\ic\,x^{1/2}} - \Psi\pars{-\ic\,x^{1/2}} \over 2\ic\,x^{1/2}}
\\[3mm]&=
-\,{1 \over x}
-
\ic\,{1 \over 2x^{1/2}}\,\bracks{%
{\Psi\pars{\ic\,x^{1/2}} - \Psi\pars{1 +\ic\,x^{1/2}}}
+
\pi\cot\pars{-\pi\ic\,x^{1/2}}}
\\[3mm]&=
-\,{1 \over x}
-
\ic\,{1 \over 2x^{1/2}}\,\bracks{%
-\,{1 \over \ic\,x^{1/2}} + \ic\,{\pi \over \tanh\pars{\pi x^{1/2}}}}
=
{\pi \over 2x^{1/2}\tanh\pars{\pi x^{1/2}}} - {1 \over 2x}
\end{align}

$$
\sum_{n = 1}^{\infty}{1 \over n^{2} + x}
=
{\pi \over 2x^{1/2}\tanh\pars{\pi x^{1/2}}} - {1 \over 2x}\,,
\qquad
x > 0
$$
$$
x \gg 1
\quad\imp\quad
\sum_{n = 1}^{\infty}{1 \over n^{2} + x}
\sim
\color{#ff0000}{\large{\pi \over 2x^{1/2}}}
$$
Also, when $x \gg 1$, this asymptotic result can be found ( Riemann Sum ) as
$\pars{~N \gg 1~}$:
$$
\sim
N\int_{1/N}^{1}{{\rm d}\xi \over \xi^{2}N^{2} + x}
=
{1 \over x^{1/2}}\bracks{%
\overbrace{\arctan\pars{N \over x^{1/2}}}^{\sim\ \pi/2}
-
\overbrace{\arctan\pars{1 \over x^{1/2}}}^{\sim\ 0}}
\sim
{\pi \over 2x^{1/2}}
$$
A: The summation over "n" leads to (-1+ x Sqrt[Pi] Coth[x Sqrt[Pi]]) / (2 x). If "x" goes to infinity, the limit is zero.
Sorry for not using LaTex; it is difficult for me since I am almost blind.
