# For which $s\in\mathbb{C}$ do we have $\sum_{n=0}^\infty{s \choose n}=2^s$?

By comparison with the binomial expansion of $$(1+x)^s$$, a sufficient condition for the formula $$\sum_{n=0}^\infty{s\choose n}=2^s$$ to hold is that $$A_s=\sum_{n=0}^\infty{s\choose n}$$ is absolutely convergent, since then the binomial expansion of $$(1+x)^s$$ is normally convergent on $$|x|\leq 1$$ and represents $$(1+x)^s$$ for $$|x|<1$$ - by continuity of both sides of the equation, we get the formula for $$|x|=1$$.

I manage to prove that $$A_s$$ is absolutely convergent for real $$s\geq 1$$. Indeed, then we have with $$N=[s]+1$$ and $$m=[s]\geq 1$$ the majorant $$\sum_{n=0}^\infty\left|{s\choose n}\right|\leq\sum_{n=0}^m\left|{s\choose n}\right|+N!\zeta(m+1)\,.$$ Further, for $$s=\frac{1}{2}$$, we have by the Wallis product that $$\left|{\frac{1}{2}\choose n}\right|\sim\frac{1}{2n\sqrt{\pi n}}$$, and we also get the absolute convergence.

For the other values of $$s$$, I am stuck. The series under consideration should be absolutely convergent for all real $$s\geq 0$$. However, for $$s=-\frac{1}{2}$$ the series $$A_s$$ is convergent but not absolutely convergent by the Leibniz criterion and according to numerical evaluation it indeed represents $$2^s$$.

Question. For which complex values $$s$$ does the formula $$A_s=2^s$$ hold?

Question'. For which complex values $$s$$ is the series $$A_s$$ convergent on the nose, not necessarily absolutely convergent?

• It should be noted that $A_s$ is divergent for $s=-1$. So, it cannot hold for all $s$. Jun 27, 2022 at 7:33
• Partial answers are welcome, especially the absolute convergence of $A_s$ for $s\geq 0$. Jun 27, 2022 at 8:35
• The numerical experiments suggest that this result is rrue for $\Re{s}>-1$. Jun 27, 2022 at 10:16

A consequence of Stirling's formula, for $$s$$ not a nonnegative integer, is $$\binom{s}{n}=\frac{(-1)^n}{n!}\frac{\Gamma(n-s)}{\Gamma(-s)}=\frac1{\Gamma(-s)}\frac{(-1)^n}{n^{s+1}}\left(1+\frac{s(s+1)}{2n}+o(1/n)\right)$$ as $$n\to\infty$$; since $$\sum_{n=1}^\infty(-1)^n n^{-s-1}$$ converges for $$\Re s>-1$$ (which can be seen after pairing the terms), the same is true for $$\sum_{n=0}^\infty\binom{s}{n}$$ (the difference is an absolutely convergent series).
Since $$(1+z)^s=\sum_{n=0}^\infty\binom{s}{n}z^n$$ holds for $$|z|<1$$ and any $$s\in\mathbb{C}$$, the ability to put $$z=1$$ here (for $$\Re s>-1$$) now follows from Abel's theorem on power series.
Another approach is to use (Taylor's theorem with integral form of the remainder) $$2^s=\sum_{k=0}^{n-1}\binom{s}{k}+n\binom{s}{n}\int_0^1(1+t)^{s-n}(1-t)^{n-1}\,dt$$ and the fact that the integral is $$O(1/n)$$, but we still need to show $$\binom{s}{n}\underset{n\to\infty}{\longrightarrow}0$$ here.
• @GEdgar: in other cases (i.e. when $\Re s\leqslant-1$) $\binom{s}{n}$ doesn't tend to $0$. (This is basically contained in the first line of the answer...) Jun 27, 2022 at 12:49
• @Christoph: Just use $\log\Gamma(z)\asymp\left(z-\frac12\right)\log z-z+\frac12\log2\pi+O(z^{-1})$ to get \begin{align*}\log\frac{\Gamma(z+a)}{z^a\Gamma(z)}&\asymp\frac{a(a-1)}{2z}+O(z^{-2})\\\log\frac{\Gamma(z+a)}{\Gamma(z+b)}&\asymp(a-b)\log z+\frac{(a-b)(a+b-1)}{2z}+O(z^{-2})\end{align*} (now put $a=-s$, $b=1$, and exponentiate). Jun 27, 2022 at 16:23