Determine if the series representation is true or not

Question

A double factorial is such that $(2n)!!=(2n)(2n-2)...(2),$ and $(2n-1)!!=(2n-1)(2n-3)...3$. So $8!!=(8)(6)(4)(2)$. More information here: https://en.wikipedia.org/wiki/Double_factorial

With this in mind, I want to know 2 things. Is it true that

$$2^{-\frac{3}{2}}=2\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(2n-1)!!}{(2n)!!}n$$

I tried to convert $2^{-\frac{3}{2}}$ into its power series representation...and failed. What I did was $2^{-\frac{3}{2}}=e^{\ln(2^{-\frac{3}{2}})}= \sum_{n=0}^{\infty}\frac{\ln(2^{-\frac{3}{2}})}{n!}$, but this is no where close to what I want. If there is a way to express these double factorials differently, please inform me of it.

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And is this true:

$$\pi = 3\sum_{n=0}^{\infty}\frac{(2n+1)!!}{(2n+1)^2(2n)!!4^n}$$

This one I really have no idea how to approach. I expanded and simplified the factorials getting $3\sum_{n=0}^{\infty}\frac{(2n-1)(2n-3)...(1)}{(2n+1)(2n)...(2)4^n}$. At most it can be simplified to $\frac{(2n-1)!!}{4^n (2n+1)!!}$. Are there special properties of these double factorials that I can use? I also vaguely remember that there is a series representation of the binomial theorem that looks similar to this. Not sure how it would help though.

Answer 1

The first series would be an instance of the binomial series $(1+z)^{-3/2}$ at $z\to1$, but it diverges because $(2n-1)!!/(2n)!!\asymp(n\pi)^{-1/2}$ as $n\to\infty$, so that the terms grow like $\sqrt{n}$ in absolute value. The second one is true, as follows from $$\arcsin x=\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!!}\frac{x^{2n+1}}{2n+1}$$ at $x=1/2$ (here $(-1)!!=0!!=1$ are assumed), which in turn is obtained by (termwise) integration of $(1-x^2)^{-1/2}$, again a binomial series.