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 $\Im(s)$).

**Final addition:**

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.