<s>[INST] How to prove that $\pi=\pi$? I am trying to prove that indeed $\pi=\pi$. More precisely, that:
$$6\sum_{n=0}^\infty \frac{(2n)!}{4^n (n!)^2 (2n+1) 2^{2n+1}}=\pi$$
Where the definition of $\pi:$
$$\pi=4\sum_{n=1}^\infty\frac{(-1)^{(n+1)}}{(2n-1)}$$
Using the epsilon delta definition, we should prove:
$$\forall \epsilon_+\exists\delta\forall k(k>\delta\rightarrow|3+6\sum_{n=1}^k \frac{(2n)!}{4^n (n!)^2 (2n+1) 2^{2n+1}}-4\sum_{n=1}^k\frac{(-1)^{(n+1)}}{(2n-1)}|<\epsilon)$$
Re-arranging the sum, it can be written as:
$$\sum_{n=1}^k\frac{(-1)^{(n+1)} 4^{n+1} (n!)^2 (2n+1) 2^{(2n+1)}-6(2n)!(2n-1)}{(4n^2-1) 4^n (n!)^2 2^{2n+1} }-3$$
But  from this point it's difficult to go on and find delta as a function of epsilon. What techniques should be used to prove the identity? Looking at a graph I think it's clear that $\delta=\text{ceil}(\frac{1} {\epsilon})$ would satisfy the criteria, but proving it is a other matter.
 [/INST] Consider the expression
\begin{eqnarray*}
\sum_{n=0}^{\infty} \binom{2n}{n} \frac{x^{2n+1}}{2n+1}.
\end{eqnarray*}
Now use 
\begin{eqnarray*}
 \frac{x^{2n+1}}{2n+1}= \int_0^x x^{2n} dx
\end{eqnarray*}
and
\begin{eqnarray*}
\sum_{n=0}^{\infty} \binom{2n}{n} x^{2n} = \frac{1}{\sqrt{1-4x^2}}.
\end{eqnarray*}
Invert the order of the plum & integral and the expression becomes 
\begin{eqnarray*}
\int_0^x \frac{1}{\sqrt{1-4x^2}} dx.
\end{eqnarray*}
This integral can be done by the substition $2x=\sin( \theta)$ to give
\begin{eqnarray*}
\frac{1}{2}\sin^{-1} (2x).
\end{eqnarray*}
Now substitute $x^2=1/8$ and we have
\begin{eqnarray*}
\sum_{n=0}^{\infty} \frac{1}{2n+1}\binom{2n}{n} \frac{1}{8^{n}}= \frac{\pi}{2 \sqrt{2}}.
\end{eqnarray*}
</s>