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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?

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

1 Answer 1

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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.

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  • $\begingroup$ Very true indeed! $\endgroup$
    – Z Ahmed
    Jun 27, 2022 at 10:17
  • $\begingroup$ What remains is to prove divergence in the other cases. $\endgroup$
    – GEdgar
    Jun 27, 2022 at 11:42
  • $\begingroup$ @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...) $\endgroup$
    – metamorphy
    Jun 27, 2022 at 12:49
  • $\begingroup$ Thank you! Can you elaborate how you apply Stirlings formula. $\endgroup$ Jun 27, 2022 at 12:59
  • $\begingroup$ @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). $\endgroup$
    – metamorphy
    Jun 27, 2022 at 16:23

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