How to prove Double Factorial as Infinity Sum? [duplicate]

During one exposition of a demonstration, the professor shows the following identity:

$$\sum_{m = 0}^{\infty} \frac{(-1)^m \binom{n}{m}}{(2m+1)} = \frac{(2n)!!}{(2n+1)!!}$$

I've been attempting some identities however with no success (such as Pascal and the definition of binomial).

obs.: $4!! = 4 \cdot2$, $6!! = 6 \cdot4 \cdot 2$, $5!! = 5 \cdot3 \cdot 1$.

Any proof would be great to clear everything.

Your infinite sum is actually finite, because $ \binom{n}{m}=0, \left(\forall m>n\right) $.

Let $ n\in\mathbb{N} $, denoting $ u_{n}=\sum\limits_{m=0}^{n}{\binom{n}{m}\frac{\left(-1\right)^{m}}{2m+1}} $, we have : \begin{aligned}u_{n}=\sum_{m=0}^{n}{\binom{n}{m}\frac{\left(-1\right)^{m}}{2m+1}}&=\sum_{m=0}^{n}{\binom{n}{m}\left(-1\right)^{m}\int_{0}^{1}{x^{2m}\,\mathrm{d}x}}\\ &=\int_{0}^{1}{\left(1-x^{2}\right)^{n}\,\mathrm{d}x}\\ &=\left[x\left(1-x^{2}\right)^{n}\right]_{0}^{1}+2n\int_{0}^{1}{x^{2}\left(1-x^{2}\right)^{n-1}\,\mathrm{d}x}\\ &=2n\int_{0}^{1}{\left(1-\left(1-x^{2}\right)\right)\left(1-x^{2}\right)^{n-1}\,\mathrm{d}x}\\ u_{n}&=2n\left(u_{n-1}-u_{n}\right)\\ \iff u_{n}&=\frac{2n}{2n+1}u_{n-1}\\&=\prod_{k=1}^{n}{\frac{2k}{2k+1}}u_{0}=\frac{\left(2n\right)!!}{\left(2n+1\right)!!}\end{aligned}