# 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?

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

Thanks.

• Related (for possible answerers): math.stackexchange.com/questions/872540/… I assume this is probably one of the things that prompted the OP's question. – apnorton Jul 22 '14 at 15:39
• Using the asymptotic series for digamma yields $\sim -\pi^2/(72a^2)$ immediately. – Antonio Vargas Jul 22 '14 at 15:54
• @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! – Olivier Oloa Jul 22 '14 at 16:04
• 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 $z\in(0,1)$. – Jack D'Aurizio Jul 22 '14 at 16:05

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: $$I(\alpha) = \frac{\pi^2}{72\alpha^2}+O\left(\frac{1}{\alpha^3}\right).$$

• 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. – Olivier Oloa Jul 22 '14 at 20:42
• @OlivierOloa: Ok, that was not so hard, just a consequence of $f(x)\geq x/12$, done. – Jack D'Aurizio Jul 22 '14 at 21:03
• It's OK for me. Thanks! – Olivier Oloa Jul 22 '14 at 21:30


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}

• You seem to have forgotten the $\pi^2$ term in the final line. – Meow Jul 23 '14 at 7:12
• @Alyosha Fixed. I'm almost sleeping ( 3 a.m. right now ). Thanks. – Felix Marin Jul 23 '14 at 7:14
• @Felix Marin Thank you for this approach. (+1) – Olivier Oloa Jul 23 '14 at 13:22
• @OlivierOloa I add the closed expression you can check it. Thanks. – Felix Marin Jul 23 '14 at 17:07
• 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) ($\alpha=e^n$). – Olivier Oloa Jul 23 '14 at 17:52