<s>[INST] Prove that $\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{8^{n}}= \sqrt{2}$ using $f(x) = \frac{1}{\sqrt{1-4x}}$ Prove that $\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{8^{n}}= \sqrt{2}$ using $f(x) = \frac{1}{\sqrt{1-4x}}$
For the past 2 days I've tried to prove this but with no results.
I've done some research but with no results.
Any ideas how can I finally solve this?
 [/INST] Us the extended binomial theorem to show that:
$$(1-x)^{-1/2} = 1 +\frac{1}{1!}\frac{1}{2}x+\frac{1}{2!}\frac{1}{2}\cdot\frac{3}{2}x^2+\frac{1}{3!}\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}x^3\dots$$
And then prove (by induction or just directly) that:
$$\frac{1}{n!}\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}...\frac{2n-1}{2} = \frac{1}{4^n}\binom{2n}{n}$$
This means that $$\sum_{n=0}^{\infty} \frac{x^n}{4^n}\binom{2n}{n} = \frac{1}{\sqrt{1-x}}$$
Then substitute $4x$ for $x$ and you get a power series for $\frac{1}{\sqrt{1-4x}}$ and then use $x=1/8$.
</s>