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Is there any chance to express the series $$S=\sum_2^{\infty}\frac{\ln(n+1)}{(n^2-1)}$$ in terms of a known function?

My idea is to start from the generalized-Euler-constant function (http://arxiv.org/abs/math/0610499): $$\gamma(z)=\sum_1^{\infty}z^{n-1}\left( \frac1n-\log \frac{n+1}{n}\right)$$ Any hint?

user280778
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  • honestly i don't know. maybe one could start from a frullani style representation of log and see what one gets out – tired Oct 16 '15 at 00:14
  • it seems that the sum can be given in form of the genrealized stieltjes constants, mentioned here http://math.stackexchange.com/questions/835862/seeking-closed-form-solution-to-sum-n-1-infty-frac-log1n1n-a?rq=1 – tired Oct 16 '15 at 12:22

2 Answers2

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The series admits a closed form in terms of the poly-Stieltjes constants.

Proposition. Let $$S=\sum_{n=2}^{\infty}\frac{\ln(n+1)}{n^2-1}.$$

Then $$ \bbox[15px,border:1px solid orange]{S=\frac{\ln 2}4 -\frac{\gamma_1}2+\frac{\gamma_1\!\left(2,0\right)}2}\tag1 $$ where $$ \begin{align} &\gamma_1 = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log n}{n}-\frac12 \log^2 \!N\right) \tag2\\ &\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac12 \log^2 \!N\right). \tag3 \end{align} $$

Proof. We have $$ \begin{align} &\sum_{n=2}^{\infty}\frac{\ln(n+1)}{n^2-1}\\ &=\frac12\lim_{N \to \infty}\left(\sum_{n=2}^N\frac{\ln{(n+1)}}{n-1}-\sum_{n=2}^N\frac{\ln{(n+1)}}{n+1}\right)\\ &=\frac12\lim_{N \to \infty}\left(\sum_{n=1}^{N-1}\frac{\ln{(n+2)}}n-\sum_{n=3}^{N+1}\frac{\ln{n}}{n}\right)\\ &=\frac12\lim_{N \to \infty}\left(\sum_{n=1}^{N-1}\frac{\ln{(n+2)}}n-\frac{\ln^2 N}2\right)-\frac12\lim_{N \to \infty}\left(\sum_{n=1}^{N+1}\frac{\ln{n}}{n}-\frac{\ln^2 N}2\right)+\frac{\ln 2}4\\ &=\frac12\gamma_1\!\left(2,0\right)-\frac12\gamma_1+\frac{\ln 2}4 \end{align} $$ as announced.

Remark. Observe that $\gamma_1(0,0)$ is just $\gamma_1$, an ordinary Stieltjes constant.

Community
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Olivier Oloa
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Not a closed form, but a numerical evaluation. $$ \begin{align} \sum_{k=2}^\infty\frac{\log(k+1)}{k^2-1} &=\frac12\sum_{k=2}^\infty\log(k+1)\left(\frac1{k-1}-\frac1{k+1}\right)\\ &=\frac12\lim_{n\to\infty}\left(\sum_{k=2}^n\frac{\log(k+1)}{k-1}-\sum_{k=2}^n\frac{\log(k+1)}{k+1}\right)\\ &=\frac12\lim_{n\to\infty}\left(\sum_{k=1}^{n-1}\frac{\log(k+2)}k-\sum_{k=3}^{n+1}\frac{\log(k)}k\right)\\ &=\frac{\log(2)}4+\frac12\sum_{k=1}^\infty\frac{\log\left(1+\frac2k\right)}k\\ \end{align} $$ Using the Euler-Maclaurin Sum Formula with an error of $O\left(\frac1{k^{23}}\right)$ and using $1000$ terms of the series, we get that $$ \begin{align} &\sum_{k=2}^\infty\frac{\log(k+1)}{k^2-1}\\ &=\frac{\log(2)}4+\frac12\sum_{k=1}^\infty\frac{\log\left(1+\frac2k\right)}k\\ &\approx 1.232969586009636405434158525257658981115253440057209139211078744 \end{align} $$

robjohn
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