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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.

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  • $\begingroup$ Related (for possible answerers): math.stackexchange.com/questions/872540/… I assume this is probably one of the things that prompted the OP's question. $\endgroup$ – apnorton Jul 22 '14 at 15:39
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    $\begingroup$ Using the asymptotic series for digamma yields $\sim -\pi^2/(72a^2)$ immediately. $\endgroup$ – Antonio Vargas Jul 22 '14 at 15:54
  • $\begingroup$ @Antonio Vargas Yes! $-\frac{\pi^{2}}{72} \frac{1}{\alpha^{2}} + \frac{\pi^{4}}{10 \: 800} \frac{1}{\alpha^{4}} +\, O\!\left(\frac{1}{\alpha^{6}}\right)$ Bernouilli numbers come ... Bravo! $\endgroup$ – Olivier Oloa Jul 22 '14 at 16:04
  • $\begingroup$ It is interesting to notice that the first term of the asymptotics is just given by the inequality $\left(\frac{1}{\log z}+\frac{1}{1-z}-\frac{1}{2}\right)\frac{z}{1-z}\leq\frac{\sqrt{z}}{12}$ when $z\in(0,1)$. $\endgroup$ – Jack D'Aurizio Jul 22 '14 at 16:05
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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: $$ I(\alpha) = \frac{\pi^2}{72\alpha^2}+O\left(\frac{1}{\alpha^3}\right).$$

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  • $\begingroup$ Dear Jack, you have in fact proved that the series is $\mathcal{O}\left(\frac{1}{\alpha^{2}}\right)$, could you please improve your formulae in order to get $\sim -\frac{\pi^{2}}{72} \frac{1}{\alpha^{2}}$, so I could accept your answer. Thanks. $\endgroup$ – Olivier Oloa Jul 22 '14 at 20:42
  • $\begingroup$ @OlivierOloa: Ok, that was not so hard, just a consequence of $f(x)\geq x/12$, done. $\endgroup$ – Jack D'Aurizio Jul 22 '14 at 21:03
  • $\begingroup$ It's OK for me. Thanks! $\endgroup$ – Olivier Oloa Jul 22 '14 at 21:30
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ 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}

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  • $\begingroup$ You seem to have forgotten the $\pi^2$ term in the final line. $\endgroup$ – Meow Jul 23 '14 at 7:12
  • $\begingroup$ @Alyosha Fixed. I'm almost sleeping ( 3 a.m. right now ). Thanks. $\endgroup$ – Felix Marin Jul 23 '14 at 7:14
  • $\begingroup$ @Felix Marin Thank you for this approach. (+1) $\endgroup$ – Olivier Oloa Jul 23 '14 at 13:22
  • $\begingroup$ @OlivierOloa I add the closed expression you can check it. Thanks. $\endgroup$ – Felix Marin Jul 23 '14 at 17:07
  • $\begingroup$ This is not an answer, it is a comment including hyperlinks. @Felix Marin OK! Thanks! Your result with your integral is also a consequence of Ramanujan's result: Theorem 1 (here)+ equation (22) (here) ($\alpha=e^n$). $\endgroup$ – Olivier Oloa Jul 23 '14 at 17:52

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