Proving $\pi=(27S-36)/(8\sqrt{3})$, where $S=\sum_{n=0}^\infty\frac{\left(\left\lfloor\frac{n}{2}\right\rfloor!\right)^2}{n!}$ I have to prove that:
$$\pi=\frac{27S-36}{8\sqrt{3}}$$
where I know that $$S=\sum_{n=0}^\infty\frac{\left(\left\lfloor\frac{n}{2}\right\rfloor!\right)^2}{n!}$$
Where do I get started?
 A: 
Assignment:
Prove:
$$\pi=\frac{27\mathcal{S}-36}{8\sqrt{3}}$$
Where:
$$\mathcal{S}:=\sum_{\text{n}\space\ge\space0}\frac{\left(\left\lfloor\frac{\text{n}}{2}\right\rfloor!\right)^2}{\text{n}!}$$
Where $\left\lfloor x\right\rfloor$ is the Floor function.


Solution:
First, let's add the sum over all odd and even numbers:
\begin{equation}
\begin{split}
\mathcal{S}&=\frac{\left(0!\right)^2}{0!}+\frac{\left(0!\right)^2}{1!}+\frac{\left(1!\right)^2}{2!}+\frac{\left(1!\right)^2}{3!}+\frac{\left(2!\right)^2}{4!}+\frac{\left(2!\right)^2}{5!}+\dots\\
\\
&=\underbrace{\sum_{\text{n}\space=\space0}^\infty\frac{\text{n}!\cdot\text{n}!}{\left(2\text{n}+1\right)!}}_\text{odd part}+\underbrace{\sum_{\text{n}\space=\space0}^\infty\frac{\text{n}!\cdot\text{n}!}{\left(2\text{n}\right)!}}_\text{even part}\\
\\
&=\sum_{\text{n}\space=\space0}^\infty\frac{\text{n}!\cdot\text{n}!}{\left(2\text{n}+1\right)!}+\sum_{\text{n}\space=\space0}^\infty\frac{\text{n}!\cdot\text{n}!}{\left(2\text{n}+1\right)!}\cdot\left(2\text{n}+1\right)
\end{split}\tag1
\end{equation}
Now, let's recall that the Gamma function is defined as:
$$\Gamma\left(\text{n}+1\right)=\text{n}!\tag2$$
$\forall\text{n}\in\mathbb{N}_0$.
We can now re-write the factorials using the Gamma function and combine the sums:
\begin{equation}
\begin{split}
\mathcal{S}&=\sum_{\text{n}\space=\space0}^\infty\frac{\Gamma\left(\text{n}+1\right)\cdot\Gamma\left(\text{n}+1\right)}{\Gamma\left(2\text{n}+2\right)}+\sum_{\text{n}\space=\space0}^\infty\frac{\Gamma\left(\text{n}+1\right)\cdot\Gamma\left(\text{n}+1\right)}{\Gamma\left(2\text{n}+2\right)}\cdot\left(2\text{n}+1\right)\\
\\
&=\sum_{\text{n}\space=\space0}^\infty\left\{\frac{\Gamma\left(\text{n}+1\right)\cdot\Gamma\left(\text{n}+1\right)}{\Gamma\left(2\text{n}+2\right)}+\frac{\Gamma\left(\text{n}+1\right)\cdot\Gamma\left(\text{n}+1\right)}{\Gamma\left(2\text{n}+2\right)}\cdot\left(2\text{n}+1\right)\right\}\\
\\
&=\sum_{\text{n}\space=\space0}^\infty\frac{\Gamma\left(\text{n}+1\right)\cdot\Gamma\left(\text{n}+1\right)}{\Gamma\left(2\text{n}+2\right)}\cdot\left(2\text{n}+2\right)
\end{split}\tag3
\end{equation}
Now, recall that the Beta function is given by:
$$\beta\left(x,\text{y}\right)=\frac{\Gamma\left(x\right)\Gamma\left(\text{y}\right)}{\Gamma\left(x+\text{y}\right)}\tag4$$
Applying this to our sum gives:
$$\mathcal{S}=\sum_{\text{n}\space=\space0}^\infty\left(2\text{n}+2\right)\beta\left(\text{n}+1,\text{n}+1\right)\tag5$$
The integral representation of the Beta function is:
$$\beta\left(x+1,\text{y}+1\right)=\int\limits_0^1 t^x\left(1-t\right)^\text{y}\space\text{d}t\tag6$$
Applying this to the sum gives:
$$\mathcal{S}=\sum_{\text{n}\space=\space0}^\infty\left(2\text{n}+2\right)\int\limits_0^1 t^\text{n}\left(1-t\right)^\text{n}\space\text{d}t=2\int\limits_0^1\sum_{\text{n}\space=\space0}^\infty\left(\text{n}+1\right)\left(t\left(1-t\right)\right)^\text{n}\space\text{d}t\tag7$$
Let's recall that:
$$\sum_{\text{k}\space=\space0}^\infty\left(\text{k}+1\right)x^\text{k}=\frac{1}{\left(1-x\right)^2}\tag8$$
This can easily be proven using the Geometric series and differentiating both sides.
Using that we can re-write $(7)$:
$$\mathcal{S}=2\int\limits_0^1\frac{1}{\left(1-t\left(1-t\right)\right)^2}\space\text{d}t\tag9$$

Using a substitution $t-\frac{1}{2}=\frac{\sqrt{3}}{2}\cdot\tan\left(\theta\right)$, it is not hard to prove that:
$$\mathcal{S}=\frac{4}{3}+\frac{8\pi\sqrt{3}}{27}\tag{10}$$
Which I let you do, and that proves your result, because:
$$\mathcal{S}=\frac{4}{3}+\frac{8\pi\sqrt{3}}{27}\space\Longleftrightarrow\space\pi=\frac{27\mathcal{S}-36}{8\sqrt{3}}\tag{11}$$

