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Probably an easy question, but I found the following identity that seems true for all $s \in \mathbb{C}$:

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

Why is this the case? Do similar identities exist for $3^s,4^s,...$?

Additional observation:

For $s \in \mathbb{N}$, only a finite sum up to $s+1$ is required to get the exact power, i.e.:

$$\sum_{n=1}^{s+1} \binom{s}{n}\big(1 - \zeta(n-s)\big)=2^s$$

and also:

$$\sum_{n=1}^{s+1} \binom{s}{n}\big(1+\frac{1}{2^{n-s}} - \zeta(n-s)\big)=3^s$$

etc.

This trick doesn't seem to work for negative integers or non-integer values of $s$.

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    Well, for $4^s$ you can square the equation. – 2017-01-20
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    Obviously :-) Let's go for the prime powers first then. – 2017-01-20
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    I assume you define the term for $s=n-1$ to be $\frac1n$ (its limit for $s\to n-1$) ? – 2017-01-21

2 Answers 2

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\begin{eqnarray*} \sum_{n=1}^{\infty} \binom{s}{n}(1- \zeta(n-s)) =\sum_{n=1}^{\infty}\sum_{m=2} ^{\infty}\binom{s}{n} \frac{1}{m^{n-s}} \end{eqnarray*} Invert the sums & use the binomial theorem \begin{eqnarray*} \sum_{m=2} ^{\infty}\sum_{n=1}^{\infty}\binom{s}{n} \frac{1}{m^{n-s}}=\sum_{m=2} ^{\infty} m^s(1-(1+\frac{1}{m^s}))=\sum_{m=2}^{\infty} m^s -(1+m)^s \end{eqnarray*} Telescopicy summy thingy & it equals $2^{s}$ ... as you state. So in order to get the sum to start at $m=3$ the original sum would need to be \begin{eqnarray*} \sum_{n=1}^{\infty} \binom{s}{n}(1 +\frac{1}{2^{n-s}}- \zeta(n-s)) =3^{s} \end{eqnarray*} Not sure where to go to from here ...

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    Many thanks Donald. If there are indeed series for each power with a certain 'rythm', then we could sum these for $-s$ and reproduce $\sum_{n} \frac{1}{n^s}$. The closest to such a nested series is mentioned here: https://en.wikipedia.org/wiki/Riemann_zeta_function#Globally_convergent_series – 2017-01-20
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    The rhythm is easy: $$\sum_{n=1}^{\infty} \binom{s}{n}\left(\sum_{m=1}^{x-1} \frac{1}{m^{n-s}}- \zeta(n-s)\right) =x^{s}$$ – 2017-01-20
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    The representation $\zeta(s)=\sum_{m\geq1}\frac{1}{m^{s}}$ holds only if $\textrm{Re}(s)>1$ so you can not use it for all $s\in\mathbb{C}.$ Another thing: are you saying that $$\sum_{m\geq2}m^{s}-(m+1)^{s}$$ is a telescopic series? – 2017-01-21
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  • For $Re(s) < 0$ : $$F(s) = -\sum_{m=2}^\infty \sum_{n=1}^\infty {s \choose n} m^{s-n} = -\sum_{m=2}^\infty m^s((1+\frac{1}{m})^s-1) = \sum_{m=2}^\infty (m^s-(m+1)^s) = 2^s $$

    Also as $N \to \infty$ : $(1+\frac{1}{m})^s-1 = \sum_{n=1}^N {s \choose n} m^{s-n}+ \mathcal{O}(m^{s-N})$ so inverting the double sum is allowed $$F(s) = -\sum_{n=1}^\infty \sum_{m=2}^\infty {s \choose n} m^{s-n} = \sum_{n=1}^\infty {s \choose n}(1-\zeta(n-s))$$

  • For any $s \not \in \mathbb{N}$, as $n \to \infty$ : $1-\zeta(n-s) = \mathcal{O}(2^{s-n})$ so that $\sum_{n=1}^\infty {s \choose n}(1-\zeta(n-s))$ converges and is analytic

$\implies$ by analytic continuation $$\sum_{n=1}^\infty {s \choose n}(1-\zeta(n-s)) = 2^s$$ is true for any $s$ not an integer

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    Many thanks for your complete answer, user1952009. – 2017-01-22
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    How do you justify the rearrangement in the first line? Because that's the interesting part. – 2017-01-28
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    @barto What do you mean ? What needs a justification is $\sum_{m=2}^\infty \sum_{n=1}^\infty {s \choose n} m^{s-n} = \sum_{n=1}^\infty\sum_{m=2}^\infty {s \choose n} m^{s-n}$ for $Re(s) < 0$ – 2017-01-28
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    Never mind; I didn't read it properly. As for the $O$-estimates; I assume they're valid but it seems you aren't being very explicit on whether they're locally uniform ;) – 2017-01-28
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    @barto I had to say something like $(1+x)^s-1 = \sum_{n=1}^N {s \choose n} x^{n}+ \mathcal{O}(x^{N})$ uniformly for $|x|\le 1-\epsilon$ small and $s\in \Omega$ compact – 2017-01-28