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Find when $\sum_{n=1}^{\infty }\binom{\alpha}{n}$ ($\alpha$ is a real number) diverges, converges, or converges absolutely.

First I notice that it is basically ${(1+1)}^{\alpha}$ so the sum always converges to $2^{\alpha}$. But what if I want to find the solution the other way for example using convergence test (but ratio test obviously doesn't work)? And how can I deal with absolute convergence there?

user186421
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    If it converges, it converges to $2^{\alpha}$. We have $\binom{-1}{n} = (-1)^n$, so the series diverges e.g. for $\alpha = -1$. – Daniel Fischer Dec 15 '15 at 15:29

1 Answers1

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Clearly the series converges absolutely when $\alpha = 0, 1, 2, \cdots$. So it is sufficient to consider only the case $\alpha \in \Bbb{C} \setminus \{0, 1, 2, \cdots \}$. Then we may write

\begin{align*} \binom{\alpha}{n} &= \frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!} \\ &= (-1)^n \frac{(-\alpha)\cdots(n-1-\alpha)}{n!} \\ &= (-1)^n \frac{\Gamma(n-\alpha)}{\Gamma(-\alpha)n!}. \end{align*}

Now by the Stirling's formula, it is easy to check that

$$ \frac{\Gamma(n-\alpha)}{n!} \sim \frac{1}{n^{\alpha+1}}. \tag{*} $$

This estimate shows that for $\alpha \in \Bbb{C} \setminus \{0, 1, 2, \cdots \}$, we have the followings:

  • If $\Re(\alpha) \leq -1$, then the series $\sum_{n=0}^{\infty} \binom{\alpha}{n}$ diverges since the general term does not converge to 0.

  • If $\Re(\alpha) > 0$, then the series $\sum_{n=0}^{\infty} \binom{\alpha}{n}$ converges absolutely by the limit comparison test.

  • If $-1 < \Re(\alpha) \leq 0$, then paring the $(2k)$-th term and the $(2k+1)$-th term improves the decay speed:

    $$ \frac{\Gamma(2k-\alpha)}{(2k)!} - \frac{\Gamma(2k+1-\alpha)}{(2k+1)!} = \frac{\alpha+1}{2k+1} \frac{\Gamma(2k-\alpha)}{(2k)!} \sim \frac{\alpha+1}{(2k)^{\alpha+2}}. $$

    Thus the series $\sum_{n=0}^{\infty} \binom{\alpha}{n}$ converges. Since this does not converge absolutely, the series converges conditionally.

Remark 1. The map $z \mapsto 1/\Gamma(z)$ is an entire function having zero exactly at $z = 0, -1, -2, \cdots$. This guarantees that $1/\Gamma(-\alpha)$ never vanishes whenever $\alpha \neq 0, 1, 2, \cdots$.

Remark 2. A weaker version of the estimate $\text{(*)}$, namely

$$ \frac{c}{n^{\Re(\alpha)+1}} \leq \left| \frac{\Gamma(n-\alpha)}{n!} \right| \leq \frac{C}{n^{\Re(\alpha)+1}} $$

for some constants $C > c > 0$ depending only on $\alpha$, can be obtained by a much elementary analysis. This weaker version is enough for our solution.

Sangchul Lee
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