TITLE: How is $ \sum_{n=1}^{\infty}\left(\psi(\alpha n)-\log(\alpha n)+\frac{1}{2\alpha n}\right)$ when $\alpha$ is great?
QUESTION [7 upvotes]: Let $\psi := \Gamma'/\Gamma$ denote the digamma function. 
Could you find, as $\alpha$ tends to $+\infty$, an equivalent term for the following series?

$$
\sum_{n=1}^{\infty}  \left( \psi (\alpha n) - \log (\alpha n) + \frac{1}{2\alpha n} \right)
$$

Please I do have an answer, I'm curious about different approaches. 
Thanks.

REPLY [4 votes]: $\newcommand{\+}{^{\dagger}}
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With Digamma Identity $\ds{\bf\mbox{6.3.21}}$
\begin{align}&\color{#66f}{\large\sum_{n = 1}^{\infty}\bracks{%
\Psi\pars{\alpha n} - \ln\pars{\alpha n} + {1 \over 2\alpha n}}}
\\[3mm]&=\sum_{n = 1}^{\infty}\bracks{-2\int_{0}^{\infty}{t\,\dd t \over \pars{t^{2} + \alpha^{2}n^{2}}\pars{\expo{2\pi t} - 1}}}\tag{1}
\\[3mm]&=-\,{2 \over \alpha^{2}}\sum_{n = 1}^{\infty}{1 \over n^{2}}
\int_{0}^{\infty}{t\,\dd t \over \bracks{1 +t^{2}/\pars{\alpha^{2}n^{2}}}
\pars{\expo{2\pi t} - 1}}
\\[3mm]&=-\,{2 \over \alpha^{2}}\sum_{n = 1}^{\infty}{1 \over n^{2}}
\int_{0}^{\infty}\pars{1 - {t^{2} \over \alpha^{2}n^{2}} + {t^{4}\over \alpha^{4}n^{4}} - \cdots}{t\,\dd t \over  \expo{2\pi t} - 1}
\\[3mm]&=\bracks{-2\,
\overbrace{\sum_{n = 1}^{\infty}{1 \over n^{2}}}^{\ds{\pi^{2} \over 6}}\
\overbrace{\int_{0}^{\infty}{t\,\dd t \over \expo{2\pi t} - 1}}^{\ds{1 \over 24}}}\
\,{1 \over \alpha^{2}}
+\bracks{2\
\overbrace{\sum_{n = 1}^{\infty}{1 \over n^{4}}}^{\ds{\pi^{4} \over 90}}\
\overbrace{\int_{0}^{\infty}{t^{3}\,\dd t \over \expo{2\pi t} - 1}}
^{\ds{1 \over 240}}}\,{1 \over \alpha^{4}}
\\[3mm]&\phantom{}+\bracks{-2\
\overbrace{\sum_{n = 1}^{\infty}{1 \over n^{6}}}^{\ds{\pi^{6} \over 945}}\
\overbrace{\int_{0}^{\infty}{t^{5}\,\dd t \over \expo{2\pi t} - 1}}
^{\ds{1 \over  504}}}\,{1 \over \alpha^{6}} + \cdots
\\[3mm]&=\color{#66f}{\large-\,{\ \pi^{2} \over 72}\,{1 \over \color{#c00000}{\alpha^{2}}} + {\ \pi^{4} \over 10800}\,{1 \over \color{#c00000}{\alpha^{4}}}
- {\ \pi^{6} \over 238140}\,{1 \over \color{#c00000}{\alpha^{6}}} + \cdots}
\end{align}

Indeed, there is a closed expression, in terms of an integral, because the serie in $\pars{1}$ is given by:
  $$
\sum_{n = 1}^{\infty}{1 \over t^{2} + \alpha^{2}n^{2}}
={1 \over 2\alpha t^{2}}\,\bracks{\pi t\coth\pars{{\pi \over \alpha}\,t} - \alpha}
$$

\begin{align}&\sum_{n = 1}^{\infty}\bracks{%
\Psi\pars{\alpha n} - \ln\pars{\alpha n} + {1 \over 2\alpha n}}
=\int_{0}^{\infty}
{1 - \pars{\pi t/\alpha}\coth\pars{\pi t/\alpha} \over t}
\,{\dd t \over \expo{2\pi t} - 1}
\end{align}