Why is $\sum_{n\ge1} \binom{s}{n}\left(1 - \zeta(n-s)\right)=2^s$ for all $s \in \mathbb{C}$?

Question

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 $s$.

Answer 1

\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 ...

Answer 2

• 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