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Calculate the following sequence $$\sum_{n=0}^{+\infty }\left ( -\dfrac{1}{4\alpha } \right )^{n}\dfrac{ (2n)!}{n!},\; \alpha >0$$

Iloveyou
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    The sequence diverges for every $\alpha$, because the individual terms go to infinity. Hint: show that $\frac{(2n)!}{n!}\geq n!$; there's an 'aha!' that should make this obvious. – Steven Stadnicki Apr 16 '14 at 05:16
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    As a series, this one diverges badly. If I'm not mistaken, this is the asymptotic expansion of $\sqrt{\pi\alpha}e^{\alpha} \text{erfc}(\sqrt{\alpha})$ where $\text{erfc}(x)$ is the complemenary error function $$\text{erfc}(x) = \frac{2}{\sqrt{\pi}}\int_x^\infty e^{-t^2} dt$$ Look at the wiki page of Error function for more details. – achille hui Apr 16 '14 at 06:23
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    @achillehui: Nice catch! Adjusting this answer yields $$\frac2{\sqrt{\pi}}\int_x^\infty e^{-t^2},\mathrm{d}t \sim\frac1{\sqrt{\pi}}e^{-x^2} \sum_{k=0}^\infty\frac{(-1)^k(2k)!}{4^kk!x^{2k+1}}$$ which transforms into your expansion with $x=\sqrt\alpha$. – robjohn Apr 16 '14 at 12:36

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Note that $\dfrac{(2n)!}{n!}=(2n)(2n-1)\cdots(n+1)\ge n^n$, therefore, $$ \left|\,\left(-\frac1{4\alpha}\right)^n\frac{(2n)!}{n!}\,\right| \ge\left|\,\frac{n}{4\alpha}\,\right|^n\tag1 $$ Thus, the terms of the series do not go to $0$, so the series diverges.

However, if as suggested by Lucian, this is supposed to be $$ \sum_{k=0}^\infty\binom{2n}{n}\left(-\frac1{4\alpha}\right)^{\large n}\tag2 $$ then $$ \begin{align} \binom{2n}{n} &=2^n\frac{(2n-1)!!}{n!}\tag{3a}\\ &=4^n\frac{\left(n-\frac12\right)!}{n!\left(-\frac12\right)!}\tag{3b}\\ &=4^n\binom{n-\frac12}{n}\tag{3c}\\[3pt] &=(-4)^n\binom{-\frac12}{n}\tag{3d} \end{align} $$ Explanation:
$\text{(3a)}$: $(2n)!=(2n-1)!!\,2^nn!$
$\text{(3b)}$: $(2n-1)!!=2^n\frac{\left(n-\frac12\right)!}{\left(-\frac12\right)!}$
$\text{(3c)}$: write ratio as a binomial coefficient$\\[9pt]$
$\text{(3d)}$: negative binomial coefficient

Thus, $$ \begin{align} \sum_{n=0}^\infty\binom{2n}{n}\left(-\frac1{4\alpha}\right)^{\large n} &=\sum_{n=0}^\infty\binom{-\frac12}{n}\frac1{\alpha^n}\tag{4a}\\[3pt] &=\left(1+\frac1\alpha\right)^{-1/2}\tag{4b}\\[6pt] &=\sqrt{\frac{\alpha}{\alpha+1}}\tag{4c} \end{align} $$

robjohn
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This is a binomial series. And also, it's $(n!)^2$.

Lucian
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  • How is this a binomial series? The 'top' of the binomial coefficient is variable, not constant, even with the correction you suggest... – Steven Stadnicki Apr 16 '14 at 05:19
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    It's the binomial series of $\dfrac1{\sqrt{1-4x}}$ , for $x=-\dfrac1{4a}$ . The central binomial coefficient stems from $-1/2\choose n$ – Lucian Apr 16 '14 at 05:24
  • Oh, of course! I'm sorry, it's late and apparently my brain is leaking. – Steven Stadnicki Apr 16 '14 at 05:26
  • If It's $(n!)^2$, then $S_N=\sum_{n=0}^{N}\left ( -\dfrac{1}{4\alpha } \right )^{n}\dfrac{ (2n)!}{(n!)^2}=(1-\frac{1}{4\alpha})^N$.

    Because $1-\frac{1}{4\alpha}<1\to S=\lim_{N\to \infty} S_N=0$

    $$\sum_{n=0}^{+\infty }\left ( -\dfrac{1}{4\alpha } \right )^{n}\dfrac{ (2n)!}{(n!)^2}=0$$

    Very easy!

     – Iloveyou Apr 16 '14 at 05:33
  • Very easy, and also very wrong, I'm afraid. ${N\choose n}\neq{2n\choose n}$ – Lucian Apr 16 '14 at 05:56