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Using the Taylor expansion for ${(1+x)}^{-1/2}$ we have ${(1+x)}^{-1/2}= \sum_{n=0}^\infty \binom{-1/2}{n} (x^n)$ for $|x|<1$.

But if $|a| <1$, how can we use the above fact to find

$\sum_{n=0}^\infty \binom{2n}{n} a^n?$

Thanks! Help much appreciated.

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    Please note that you did not give the expansion of $\sqrt{1+x}$, but of $(1+x)^{-1/2}$. If you want to **start** from the expansion of $(1+x)^{1/2}$, which is what you wrote except it is $\binom{1/2}{n}$, **not** $\binom{-1/2}{n}$, first differentiate term by term to get the expansion of $(1/2)x^{-1/2}$, then multiply both sides by $2$.2011-11-15

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Write out what $\binom{-1/2}{n}$ means; i.e., $ \begin{align} \binom{-1/2}{n} &= \frac{(-1/2)(-3/2) \cdots ((-2n+1)/2)}{n!} = \frac{(-1)^n}{2^n} \frac{(1)(3) \cdots (2n-1)}{n!} \\ &= \frac{(-1)^n}{2^n} \frac{(2n)!}{2(4) \cdots (2n)n!} = \left(\frac{-1}{4}\right)^n \frac{(2n)!}{n!n!} \\ &= \left(\frac{-1}{4}\right)^n \binom{2n}{n}. \end{align} $ This should be enough for you to be able to find $\sum_{n=0}^{\infty} \binom{2n}{n} a^n$.


Are you sure you mean $\sqrt{1+x}$, though? That would give $\sqrt{1+x} = \sum_{n=0}^{\infty} \binom{1/2}{n} x^n$. Then, following through the same argument as above you would obtain $\binom{1/2}{n} = \frac{-1}{2n-1} \left(\frac{-1}{4}\right)^n \binom{2n}{n}$, which would be a bit more difficult to deal with because of the $2n-1$ in the denominator.

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    You're right! It was (1+x)^(-1/2). Thanks so much for the help! I'll edit the post to reflect changes, to help others find this problem.2011-11-15