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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?

metamorphy
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Christoph Mark
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1 Answers1

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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.

metamorphy
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  • Very true indeed! – Z Ahmed Jun 27 '22 at 10:17
  • What remains is to prove divergence in the other cases. – GEdgar Jun 27 '22 at 11:42
  • @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...) – metamorphy Jun 27 '22 at 12:49
  • Thank you! Can you elaborate how you apply Stirlings formula. – Christoph Mark Jun 27 '22 at 12:59
  • @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). – metamorphy Jun 27 '22 at 16:23
  • Thank you! For $s\geq 0$ where the series is even absolutely convergent, there must be an elementary solution without Gammafunctions and Abel's Theorem (this is redundant because we know that the exponential series is then normally convergent for $|x|\leq 1$ and hence continuous in $1^-$) as this is Exercise 10 in Chapter 8 of Königsberger's Analysis 1. Stirling's formula for discrete $n$ is however developed in Chapter 8 with error term $\vartheta_n/12n$, $\vartheta_n\in(0,1]$. If you know an elementary solution, feel free to repost. Your help is appreciated! – Christoph Mark Jun 28 '22 at 10:25
  • I see, great. OK. – Christoph Mark Jun 28 '22 at 11:07