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I'm looking for a closed-form solution to this infinite series:

$$S(\alpha):=\sum_{n=1}^{\infty}\frac{\log{(1+n)}}{(1+n)^{\alpha}-1},~~~\Re(\alpha)>1.$$


My attempt

All I've really been able to do is confirm this series converges for the specified values of $\alpha$ via the integral test:

$$\begin{align} S(\alpha)\leq I(\alpha)&=\int_{0}^{\infty}\frac{\log{(1+x)}}{(1+x)^{\alpha}-1}\mathrm{d}x\\ &=\int_{0}^{\infty}\frac{u\,e^u}{e^{\alpha u}-1}\mathrm{d}u\\ &=\int_{0}^{\infty}ue^u\sum_{n=1}^{\infty}e^{-n\alpha u}\mathrm{d}u\\ &=\sum_{n=1}^{\infty}\int_{0}^{\infty}ue^ue^{-n\alpha u}\mathrm{d}u\\ &=\sum_{n=1}^{\infty}\frac{1}{(n\alpha-1)^2},~~~\text{for }\Re(\alpha)>1\\ &=\frac{1}{\alpha^2}\psi^{(1)}{\left(1-\frac{1}{\alpha}\right)} \end{align}$$

where $\psi^{(1)}(x)$ is the first derivative of the digamma function.

I thought about expanding the logarithmic part in terms of its power series to write $S(\alpha)$ as a double-sum, and then try switching the order of summation, e.g.:

$$\begin{align} S(\alpha)&=\sum_{n=1}^{\infty}\frac{\log{(1+n)}}{(1+n)^{\alpha}-1}\\ &=-\sum_{n=1}^{\infty}\frac{1}{(1+n)^{\alpha}-1}\sum_{k=1}^{\infty}\frac{(-1)^kn^k}{k}\\ &=-\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\frac{n^k}{(1+n)^{\alpha}-1}\\ &=-\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\sum_{n=1}^{\infty}\frac{n^k}{(1+n)^{\alpha}-1}. \end{align}$$

But I'm not sure how to proceed from there. Suggestions?

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  • 4
    $\begingroup$ I would be astonished if this sum had a closed form. I'd be surprised if $S(2)$ had a closed form. $\endgroup$ – Gerry Myerson Jun 16 '14 at 9:32
  • $\begingroup$ @David H You can not expand $\log (1+n)$ as $\sum_{k=1}^{\infty}\frac{(-1)^{k-1}n^k}{k}$ as $n>1$. The radius of convergence of the power series for $\log (1+x)$ is equal to $1$. $\endgroup$ – Olivier Oloa Sep 5 '15 at 20:41
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I'd be surprised if $S(2)$ had a closed form.

$S(2)$ admits a closed form in terms of the poly-Stieltjes constants unveiled here.

Moreover, $S(3),S(4),S(5),\cdots$, admit a closed form in terms of these special functions.

We have the following result.

Proposition. Let $\alpha=2,3,4,5,\ldots$.

Then $$ \bbox[15px,border:1px solid orange]{S(\alpha)=\frac1\alpha\sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}\:\gamma_1\!\!\left(1,1-e^{2k\pi i/\alpha}\right)} \tag1 $$ where $$ \gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac12 \log^2 \!N\right). \tag2 $$

Proof. For $\alpha=2,3,\ldots $, we have $$ \begin{align} S(\alpha):&=\lim_{N \to \infty}\sum_{n=1}^N\frac{\log{(1+n)}}{(1+n)^{\alpha}-1}\\ &=\frac1\alpha\lim_{N \to \infty}\left(\sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}\sum_{n=1}^N\frac{\log (n+1)}{n+1-e^{2k\pi i/\alpha}}\right)\\ &=\frac1\alpha\lim_{N \to \infty}\left(\sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}\left(\sum_{n=1}^N\frac{\log (n+1)}{n+1-e^{2k\pi i/\alpha}}-\frac12\log^2 N\right)+\frac12\log^2 N\sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}\right)\\ &=\frac1\alpha\lim_{N \to \infty}\left(\sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}\left(\sum_{n=1}^N\frac{\log (n+1)}{n+1-e^{2k\pi i/\alpha}}-\frac12\log^2 N\right)+0\right)\\ &=\frac1\alpha\sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}\lim_{N \to \infty}\left(\sum_{n=1}^N\frac{\log (n+1)}{n+1-e^{2k\pi i/\alpha}}-\frac12\log^2 N\right)\\ &=\frac1\alpha\sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}\gamma_1(1,1-e^{2k\pi i/\alpha}) \end{align} $$ where we have used a partial fraction decomposition over the complex numbers, $$ \frac{1}{X^\alpha-1}=\frac1\alpha\sum_{k=0}^{\alpha-1}\frac{e^{2k\pi i/\alpha}}{X-e^{2k\pi i/\alpha}},\quad \alpha=1,2,3,\ldots, \tag3 $$ giving $$ \frac{\log (n+1)}{(n+1)^\alpha-1}=\frac1\alpha\sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}\frac{\log (n+1)}{n+1-e^{2k\pi i/\alpha}},\quad \alpha=1,2,3,\ldots, $$ and we have used the standard fact that $\displaystyle \sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}=0$.

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  • $\begingroup$ Thank you very much for this response! I remember I had a very specific reason for asking it at the time (presumably it provided a way to solve some other difficult problem), but alas I've completely forgotten what I wanted to use it for. Out of curiosity, what sparked you to address this problem after all this time? $\endgroup$ – David H Sep 8 '15 at 20:28
  • $\begingroup$ You are welcome! I just stumbled upon your question. Thanks. @DavidH $\endgroup$ – Olivier Oloa Sep 9 '15 at 5:29

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