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How do you prove that $\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}$?
I tried to identify the sum as a binomial series, but the $4$ and the $-1/2$ puzzle me.
(This series arises in studying the first passage time of a simple random walk.)

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See Mike Spivey's comment here:… – Byron Schmuland Oct 2 '11 at 15:58
@Byron: Thanks, this answer my question. – Nicolas Essis-Breton Oct 2 '11 at 16:03
If it were the usual sort of binomial series then the $2n$ that sits in the top position in $\dbinom{2n}{n}$ would be something that does not change as the $n$ in the expression $\sum\limits_{n=0}^\infty$ goes from $0$ to $\infty$. – Michael Hardy Oct 2 '11 at 17:33
This answer of mine is related to your question. – Did Oct 5 '11 at 8:54
up vote 6 down vote accepted

The key identities are the duplication formula for the factorial (which I'll recast in a more convenient format):

$$\binom{2n}{n}=\frac{4^n}{\sqrt \pi}\frac{\left(n-\frac12\right)!}{n!}$$

and the reflection formula


Making the appropriate replacements, we obtain

$$\binom{2n}{n}=(-4)^n\frac{\sqrt \pi}{n!\left(-n-\frac12\right)!}=(-4)^n\frac{\left(-\frac12\right)!}{n!\left(-n-\frac12\right)!}=(-4)^n\binom{-\frac12}{n}$$

You can proceed from that...

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It seems that the trick doesn't work on $\sum_n\binom{3n}nz^n$. – Frank Science Aug 8 '12 at 14:46
I think they are unrelated. Maybe Zeilberger-Gosper's algorithm on CMath works. I have not verified. – Frank Science Aug 8 '12 at 14:53
Hmm, yes, I was hasty there. There's the hypergeometric route, of course. – J. M. Aug 8 '12 at 15:04

Or, by definition. \begin{eqnarray*} {-1/2\choose n}&=&{(-1/2)(-1/2-1)(-1/2-2)\cdots(-1/2-[n-1])\over n!}\cr &=&{(-1)^n\over 2^n} {(1)(3)(5)\cdots(2n-1)\over n!}\cr &=&{(-1)^n\over 2^n} {(1)(3)(5)\cdots(2n-1)\over n!}\cdot{2^n n!\over 2^n n!}\cr &=&{(-1)^n\over 4^n} {2n\choose n}. \end{eqnarray*}

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