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With $\gamma$ being the Euler Mascheroni constant, this series is well known:

$$1- \sum_{n=2}^{\infty} \frac{\zeta(n)-1}{n} = \gamma \tag{1}$$

The following series involving $\zeta(2n+1)$ also seems to converge to the same value, albeit slower:

$$1- \sum_{n=1}^{\infty} \frac{\zeta(2n+1)}{(n+1)\,(2n+1)} = \gamma \tag{2}$$

Is there a way to derive (2) from (1) ?

Robert Z
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Agno
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  • I have a very nice proof, but it is too long to write it here as a comment! On a more serious note, see https://math.stackexchange.com/questions/360395/a-tough-series-sum-k-1-infty-frac-zeta2k1-1k1-gamma-log2?rq=1 which tackles a similar problem. – Hans Olo Jul 20 '21 at 15:59
  • Perhaps $\sum\limits_{n=1}^{\infty }\frac{\zeta(2 n)-1}{2 n}=\frac{\log (2)}{2}$ provides some insight. – Steven Clark Jul 20 '21 at 16:02
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    @Hans Olo I remember that same apology was used in France a few hundred years ago ;-) Thanks for the link, I will explore this further and already noted the Harmonic numbers (that I expect to appear somewhere). – Agno Jul 20 '21 at 16:03

1 Answers1

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Note that $$\frac{\zeta(2n+1)}{(n+1)\,(2n+1)}=2\frac{\zeta(2n+1)-1}{2n+1}-\frac{\zeta(2n+1)-1}{n+1}+\frac{2}{2n+1}-\frac{2}{2n+2}.$$ Hence $$\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(n+1)\,(2n+1)}= 2\sum_{n=1}^{\infty}\frac{\zeta(2n+1)-1}{2n+1}-\sum_{n=1}^{\infty}\frac{\zeta(2n+1)-1}{n+1}+2\sum_{n=3}^{\infty}\frac{(-1)^{n-1}}{n}.$$ Finally we apply Question about series involving zeta function? and A Tough Series $\sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1}=-\gamma+\log(2)$ : $$\begin{align}\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(n+1)\,(2n+1)}&= 2\left(1-\gamma-\frac{\log(2)}{2}\right)-\left(-\gamma+\log(2)\right)+2\left(\log(2)-1+\frac{1}{2}\right)\\ &=\left(2-2\log(2)-\gamma\right)+2\log(2)-1\\ &= 1-\gamma.\end{align}$$ For a selfcontained proof, just note that
$$\begin{align}2\sum_{n=1}^{\infty}&\frac{\zeta(2n+1)-1}{2n+1}-\sum_{n=1}^{\infty}\frac{\zeta(2n+1)-1}{n+1}\\ &=\sum_{k=2}^\infty \sum_{n=1}^\infty \frac{2}{(2n+1)k^{2n+1}}-\sum_{k=2}^\infty\sum_{n=1}^\infty\frac{1}{(n+1)k^{2n+1}}\\ &=\sum_{k=2}^\infty \left(-\log \left(1-\frac{1}{k}\right)+\log \left( 1+\frac{1}{k}\right)-\frac{2}{k}\right)\\ &\qquad +\sum_{k=2}^\infty \left(\frac{1}{k}+ k\log \left( 1+\frac{1}{k}\right)+ k\log \left( 1-\frac{1}{k}\right)\right)\\ &=\sum_{k=2}^\infty \left((k+1)\log \left( 1+\frac{1}{k}\right) +(k-1)\log \left( 1-\frac{1}{k}\right)-\frac{1}{k}\right)\\ &=\sum_{k=2}^\infty \left((k+1)\log \left(k+1\right)-2k\log(k) +(k-1)\log \left( k-1\right)-\frac{1}{k}\right)\\ &=\lim_{N\to \infty}\left((N+1)\log \left(N+1\right) -2\log(2)-N\log(N)-H_N+1\right)\\ &=\lim_{N\to \infty}\left((N+1)\log \left(N+1\right) -2\log(2)-N\log(N)-\log(N)-\gamma+1\right)\\ &=\lim_{N\to \infty}\left((N+1)\log \left(1+\frac{1}{N}\right) -2\log(2)-\gamma+1\right)\\ &=2-2\log(2)-\gamma. \end{align}$$

Robert Z
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