# Does this infinite sum provide a new analytic continuation for $\zeta(s)$?

It is well known that the infinite sum:

$$\displaystyle \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$$

only converges for $\Re(s)>1$.

The Dirichlet 'alternating' sum:

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

allows for analytic continuation towards $\Re(s)>0$, but also does introduce additional zeros.

There are also nested sums see formulas 21 and 22 here that fully extend $\zeta(s)$ towards $s \in \mathbb{C}/1$.

However, I found this surprisingly simple infinite sum that Maple evaluates correctly for $\Re(s) >-1$:

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

Following Peter's suggestion below, it can be easily shown that by splitting, re-indexing via $n \mapsto n-1$ and then recombining the sums, that for $\Re(s) > 1$ the formula correctly reduces to $\displaystyle \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$.

For the analytically continued domain $-1<\Re(s)<1$, the operations of re-indexing and recombining the split sums are no longer allowed, since the sums are now diverging.

I played a bit with the individual terms in the infinite sum and it is easy to see that for $s=0$ and $s=1$ all terms will be zero. Therefore $\zeta(0)$ will be fully determined by $\frac{s+1}{2\,(s-1)} = \frac{1}{-2}=-\frac12$. When $s\rightarrow 1$, there will be a pole induced by $\frac{s+1}{2(s-1)}$, however the pole $\frac{1}{2(s-1)}$ will be offset by the infinite sum approaching $0$ and converges to the nice result $\gamma -\frac12$ with $\gamma$ being the Euler-Mascheroni constant.

Did not have much time to go deeper, but did find a nice result for $s=2$ (n=1..7):

$s=2 \rightarrow \frac{1 }{4}+ \frac{1 }{36}+\frac{1}{144}+\frac{1}{400}+\frac{1}{900}+\frac{1}{1764}+\frac{1}{3136}+\dots$ = sum of 1/A035287

For $s=3$ the following pattern emerges, however I did not find a good link to Sloane's.

$s=3 \rightarrow \frac{3 }{4}+\frac{5 }{108}+\frac{7}{864}+\frac{9}{4000}+\frac{11}{13500}+ \frac{13}{37044}+\frac{15}{87808}+\dots$

Questions:

(1) Has anyone seen this formula before? If so, I'd be grateful for a reference.

(2) To avoid this is just Maple playing a trick on me here, I am curious to learn whether the results can be reproduced in other CAS. (EDIT: now confirmed in Maple, Sage, Pari/GP and Mathematica that the formula works correctly, although numerical discrepancies are found for higher values $> 10^8$ of $\Im(s)$).

Paul Garrett´s reference below gives a good method to construe the formula above. However, I realized myself that I posted the formula on MO without any explanation on how I did find it in the first place. So, here it is to make the question complete before it vanishes into MO´s history:

Assume $s \in \mathbb{C}$, $\Re(s) \ge 0$ and take the well known expression:

$$\zeta(s) = \dfrac{s}{s-1} - \frac12+s \int_1^\infty \frac{1/2-\{x\}}{x^{s+1}}\,\mathrm{d}x$$

and substitute the fractional part of $\{x\}$ by a closed form (see here):

$$\displaystyle \{x\} = x - \lfloor x \rfloor = -\frac{\arctan\left(\cot\left(\pi x\right)\right)}{\pi} + \frac{1}{2}$$

which gives:

$$\displaystyle Z(s) = \dfrac{s}{s-1} - \frac12 +s \int_1^\infty \frac{\arctan\left(\cot\left(\pi x\right)\right)}{\pi x^{s+1}}\mathrm{d}x$$

For all rational fractions $-1<\frac{k}{n}<1$, Maple consistently returns closed forms with an infinite sum e.g.:

$$Z \left(\frac23\right) =\displaystyle 1-\frac{11}{4}\,\sqrt [3]{2}+\frac{2}{3}\,\sum _{{\it n}=2}^{\infty }-\frac{3}{4}\,{\frac {- \left( {\it n}+1 \right) ^{2/3}-6\, \left( {\it n}+1 \right) ^ {2/3}{\it n}+5\,{{\it n}}^{2/3}+6\,{{\it n}}^{5/3}}{{{\it n}}^{2/3} \left( {\it n}+1 \right) ^{2/3}}}$$

After experimenting with a few rationals, I saw a clear pattern emerge that I then could easily generalize into the formula above. To my surprise it also worked for irrational values and the domain $\Re(s) \ge 1$. Not sure whether Maple uses a similar technique as Paul Garrett´s suggestion below to transform the integral of the fractional parts into these infinite sums.

This can be construed as an application of Euler-Maclaurin summation, as carried out to an arbitrary number of stages in Appendix B of Montgomery-Vaughan's "Multiplicative Number Theory I: Classical Theory".

The general idea is to subtract the integral corresponding to a sum, breaking the integral into integrals over subintervals, as in the well-known idea $$\zeta(s) - {1\over s-1} \;=\; \sum_n \Big({1\over n^s} - \int_n^{n+1}{dx\over x^s}\Big) \;=\; \sum_n \Big({1\over n^s} - {n\over (s-1)n^s} + {n+1\over (s-1)(n+1)^s}\Big)$$ Elementary estimates show that the new sum is absolutely and uniformly-on-compacts convergent on $\Re(s)>0$. One can repeat the approximation of sum by integral as many times as desired, although organization becomes an issue. The sum in the question is equivalent to application of this idea one more stage. Montgomery-Vaughan show how to systematize this and express an arbitrary (finite) number of iterates in terms of Bernoulli polynomials/numbers.

(This wouldn't fit in the comment box.) It looks like your expression is just a rewritten form of the first sum. If you split your sum into two sums, reindex the second by $n \mapsto n-1$, and recombine the sums, then things cancel and you get $\sum_{n=3}^{\infty}\frac{1}{n^s}$. Then it looks like the first few terms that are left over are the $n=1$ and $n=2$ terms in the original sum. Did I miss something?

• Hm, If it is really rewritten form, why the first sum converges for Re(s)>1 and the second for Re(s)> -1 ? Shouldn't they converge in the same range if they are the same sum? – joro Aug 24 '13 at 6:07
• Peter, your operation of re-indexing and recombining the split sums into $\sum_{n=3}^{\infty}\frac{1}{n^s}$ is only allowed for $\Re(s)>1$. Note that my sum is analytically continued towards $\Re(s)>−1$, and you can't therefore simply re-index and recombine the sums in the domain $−1<\Re(s)<1$ since they diverge. – Agno Aug 25 '13 at 15:06
• The "identity principle" for holomorphic functions entails that if two functions agree in a (non-empty) right half-plane, they agree wherever analytically continued. – paul garrett Aug 26 '13 at 13:23

Some observations.

Your formula at $s=1/2$ equals the definite integral in this question

Assuming the formula is correct, it is interesting to me if it can be used to compute zeta.

According to my tests, Maple can't compute it for $s=1/2 + 10^4 i$.

Sage computes very good approximations for $\Im(s)$ small, though for $s = 1/2 + 10^8 i$ the error is large for all methods of infinite summations.