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I like to expand on this (unanswered) MSE question.

Take the following, nicely symmetrical, telescoping series for $\zeta(s)$:

$$\displaystyle \zeta(s) = \frac{1}{2\,(s-1)} \left(1+\sum _{n=1}^{\infty } \left( {\frac {n+1}{(n+1)^{s}}} - \frac{2\,(1-s)}{n^s}-{\frac {n-1}{\left( n-1 \right) ^{s}}}\right) \right), \qquad -1<\Re(s)<1$$

Now remove the $1$-term by subtracting $\sum _{n=1}^{\infty } \left( {\frac {1}{(n+1)^{s}}} - \frac{1}{n^s}\right)=-1$ from the series and we get the following, a bit less symmetrical, series that now also only converges in the critical strip:

$$\displaystyle \zeta(s) = \frac{1}{2\,(s-1)} \left(\sum _{n=1}^{\infty } \left( {\frac {n}{(n+1)^{s}}} - \frac{1-2\,s}{n^s} - {\frac {n-1}{\left( n-1 \right) ^{s}}}\right) \right), \qquad 0<\Re(s)<1$$

The numerator of the middle term is the only one dependent on $s$ and interestingly its real part only becomes $0$ when $\Re(s)=\frac12$. So, the numerator in the middle being purely imaginary seems to be a critical factor for exactly balancing the sum of the three divergent series to zero (assuming the RH).

I would appreciate any suggestions/tips on how to better assess the impact of the middle term numerator on the total sum. I have not gotten much further than taking $s=\frac12$ and getting this simple expression: $$\displaystyle \zeta\left(\frac12\right) = \sum _{n=1}^{\infty } \left(\sqrt{n-1} -{\frac {n}{\sqrt{n+1}}}\right)$$.

Thanks.

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  • $\begingroup$ What is the convention for $\frac0{0^s}$? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Aug 23, 2016 at 9:55
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    $\begingroup$ I assumed that $0^{1-s}$ is $0$ for all $s < 1$ (and this nicely fits the domains of convergence I need in both series above). For $s > 1$ is would induce division by zero and $s=1$ could be considered to be $1$ (but I don't need these values). $\endgroup$
    – Agno
    Commented Aug 23, 2016 at 12:40

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