<s>[INST] Proving $\frac{1}{\sqrt{2}}=1-\frac{1}{2^2}-\frac{1}{2!2^4}-\frac{3!!}{3!2^6}-\frac{5!!}{4!2^8}-\cdots$ How can I prove
$$\frac{1}{\sqrt{2}}=1-\frac{1}{2^2}-\frac{1}{2!2^4}-\frac{3!!}{3!2^6}-\frac{5!!}{4!2^8}-\cdots$$
I wanted to prove it by using the Taylor series of $\sqrt{2}$, but I couldnt do.
 [/INST] We have to prove:
$$\frac{1}{\sqrt{2}}=\frac{3}{4}-\sum_{n\geq 1}\frac{(2n-1)!!}{(n+1)!2^{2n+2}}=\frac{3}{4}-\frac{1}{4}\sum_{n\geq 1}\frac{(2n)!}{n!(n+1)!8^{n}}$$
that follows from:
$$ \sum_{n\geq 1}\frac{(2n)!}{n!(n+1)!}z^n = \frac{1-\sqrt{1-4z}-2z}{2z}.\tag{1}$$
$(1)$ is just a minor variation of the generating function for Catalan numbers.
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