How is $ \sum_{n=1}^{\infty}\left(\psi(\alpha n)-\log(\alpha n)+\frac{1}{2\alpha n}\right)$ when $\alpha$ is great? 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. 
 A: $\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}
A: Due to the Gauss formula:
$$-\psi(z)+\log(z) = \int_{0}^{1}\left(\frac{1}{\log u}+\frac{1}{1-u}\right)u^{z-1}\,du$$
your series is just $-I(\alpha)$ due to the dominated convergence theorem, where:
$$ I(\alpha) = \int_{0}^{1}\left(\frac{1}{\log z}+\frac{1}{1-z}-\frac{1}{2}\right)\frac{z^{\alpha-1}}{1-z^{\alpha}}\,dz.$$
Now, just like in this other question, we have that:
$$ f(z) = \left(\frac{1}{\log z}+\frac{1}{1-z}-\frac{1}{2}\right)\frac{z}{1-z} $$
is a positive, increasing and bounded function on $(0,1)$ that satisfies $f(z)\leq\frac{\sqrt{z}}{12}$. This gives that
$$ 0 \leq I(\alpha) \leq \frac{1}{6}-\frac{\pi}{12\alpha}\cot\frac{\pi}{2\alpha}=\frac{\pi^2}{72\alpha^2}+O\left(\frac{1}{\alpha^4}\right), $$
hence the limit when $\alpha$ approaches $+\infty$ is simply zero.
Moreover, since we have $f(z)\geq\frac{z}{12}$,
$$ I(\alpha)\geq\frac{\alpha+\gamma+\psi(\alpha)}{12\alpha}=\frac{\pi^2}{72\alpha^2}+O\left(\frac{1}{\alpha^3}\right),$$
hence:
$$ I(\alpha) = \frac{\pi^2}{72\alpha^2}+O\left(\frac{1}{\alpha^3}\right).$$
