Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
4  
See Mike Spivey's comment here: math.stackexchange.com/questions/37971/… –  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
1  
This answer of mine is related to your question. –  Did Oct 5 '11 at 8:54
add comment

2 Answers

up vote 5 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

$$\left(-n-\frac12\right)!\left(n-\frac12\right)!=(-1)^n\pi$$

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...

share|improve this answer
    
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
add comment

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*}

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.