# Using the Taylor expansion for ${(1+x)}^{-1/2}$, evaluate $\sum_{n=0}^\infty \binom{2n}{n} a^n$

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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Hint: ${-\frac{1}{2}\choose n} = (-\frac{1}{4})^n {2n \choose n}$ – Thomas Andrews Nov 15 '11 at 19:29
A part of this post is devoted to an answer. – Did Nov 15 '11 at 19:30
You may not quite have the right expansion. But when you get the right one, observe that for example $(-1/2)(-3/2)(-5/2)(-7/2)=(-1)^4(1/2)^4 (8!)/(2\cdot 4\cdot 6\cdot 8)$. After some simplification this is $(-1)^4(1/2^4)(1/2^4)(8!)/4!$. For the coefficient of $x^4$ in the expansion of the right thing, there is another factor of $4!$ at the bottom. – André Nicolas Nov 15 '11 at 19:33
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 $2$. – André Nicolas Nov 15 '11 at 21:11

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.