Hi I am trying to get a smart way to connect this two well known expressions for $\pi$:
Leibniz serie:
$$\frac{\pi}{4}=\sum_{n=1}^{\infty}{\frac{(-1)^{n+1}}{2n-1}}$$
Wallis product:
$$\frac{\pi}{2}=\prod_{n=1}^{\infty}{\frac{(2n)^2}{(2n-1)(2n+1)}}$$
and only saw this method through Brouckner's continued fraction:
https://link.springer.com/article/10.1007/BF00348349
It seems a little bit too tricky to be the simplest way to stablish such a connection. Any advice will be valuable.
A few ideas (may be)
A small typo in the Wallis product $$\frac{\pi}{\color{red}{2}}=\prod_{n=1}^{\infty}{\frac{(2n)^2}{(2n-1)(2n+1)}}$$
Now, we could think from $$\frac{(2n)^2}{(2n-1)(2n+1)}=1-\frac{1}{2 (2 n+1)}+\frac{1}{2 (2 n-1)}$$ and consider the logarithm and then some Taylor series after some value of $n$