Relationship between Catalan's constant and $\pi$

\begin{align}\sum_{n=0}^\infty \frac{\binom{2n}{n}^2}{4^{2n+1}(2n+1)}=\frac{\text{G}}{\pi}\tag1\end{align}

(see p81, Deriving Forsyth-Glaisher type series for $\frac{1}{\pi}$ and Catalan's constant by an elementary method. )

From the same source,

\begin{align}\sum_{n=0}^\infty \frac{\binom{2n}{n}^2}{16^n(2n+3)}=\frac{\text{G}}{\pi}+\frac{1}{2\pi}\tag2\end{align}

ADDENDUM:

Proof for (1),

It is well known that for $n\geq 0$ integer,

\begin{align}\int_0^{\frac{\pi}{2}}\cos^{2n} x\,dx=\frac{\pi}{2}\cdot\frac{\binom{2n}{n}}{4^n}\end{align}

(Wallis formula)

Therefore for $n\geq 0$ integer,

\begin{align}\frac{\binom{2n}{n}^2\pi^2}{4^{2n+1}(2n+1)}=\int_0^1 \left(\int_0^\infty \int_0^\infty t^{2n}\cos^{2n}x \cos^{2n}y \,dx\,dy \right)\,dt\end{align}

therefore,

\begin{align}\pi^2\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^2}{4^{2n+1}(2n+1)}&=\sum_{n=0}^{\infty}\left(\int_0^1 \left(\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} t^{2n}\cos^{2n}x \cos^{2n}y \,dx\,dy \right)\,dt\right)\\ &=\int_0^1 \left(\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \left(\sum_{n=0}^{\infty}t^{2n}\cos^{2n}x \cos^{2n}y\right) \,dx\,dy \right)\,dt\\ &=\int_0^1 \left(\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \frac{1}{1-t^2\cos^2 x\cos^2 y}\,dx\,dy \right)\,dt\\ \end{align}

Perform the change of variable $u=\tan x$,$v=\tan y$,

\begin{align}\pi^2\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^2}{4^{2n+1}(2n+1)}&= \int_0^1 \left(\int_0^{\infty} \int_0^{\infty}\frac{1}{(1+u^2)(1+v^2)-t^2}\,du\,dv \right)\,dt\\ &=\int_0^1 \left(\int_0^\infty \frac{1}{\sqrt{1+v^2}}\left[\frac{\arctan\left(\frac{u\sqrt{1+v^2}}{\sqrt{1+v^2-t^2}}\right)}{\sqrt{1+v^2-t^2}}\right]_{u=0}^{u=\infty}\,dv\right)\,dt\\ &=\frac{\pi}{2}\int_0^1 \left(\int_0^\infty \frac{1}{\sqrt{1+v^2}\sqrt{1+v^2-t^2}}\,dv\right)\,dt\\ &=\frac{\pi}{2}\int_0^\infty \frac{1}{\sqrt{1+v^2}}\left[\arctan\left(\frac{t}{\sqrt{1+v^2-t^2}}\right)\right]_{t=0}^{t=1}\,dv\\ &=\frac{\pi}{2}\int_0^\infty \frac{\arctan\left(\frac{1}{v}\right)}{\sqrt{1+v^2}}\,dv\\ \end{align}

Perform the change of variable $y=\dfrac{1}{x}$,

\begin{align}\pi^2\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^2}{4^{2n+1}(2n+1)}&=\frac{\pi}{2}\int_0^\infty \frac{\arctan x}{x\sqrt{1+x^2}}\,dx\\ \end{align}

Perform the change of variable $y=\arctan x$,

\begin{align}\pi^2\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^2}{4^{2n+1}(2n+1)}&=\frac{\pi}{2}\int_0^{\frac{\pi}{2}}\frac{x}{\sin x} \,dx\\ &=\frac{\pi}{2}\Big[x\ln\left(\tan\left(\frac{x}{2}\right)\right)\Big]_0^{\frac{\pi}{2}}-\frac{\pi}{2}\int_0^{\frac{\pi}{2}}\ln\left(\tan\left(\frac{x}{2}\right)\right)\,dx\\ &=-\frac{\pi}{2}\int_0^{\frac{\pi}{2}}\ln\left(\tan\left(\frac{x}{2}\right)\right)\,dx\\ \end{align}

Perform the change of variable $y=\frac{x}{2}$,

\begin{align}\pi^2\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^2}{4^{2n+1}(2n+1)}&= -\pi\int_0^{\frac{\pi}{4}}\ln(\tan x)\,dx\\ &=\pi\times \text{G}\\ \end{align}

Therefore,

\begin{align}\boxed{\sum_{n=0}^\infty \frac{\binom{2n}{n}^2}{4^{2n+1}(2n+1)}=\frac{\text{G}}{\pi}}\end{align}

Let us give a self-contained proof of Ramanujan's identity $$\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{1}{2n+1}=\frac{4G}{\pi}.\tag{1}$$ We may recall the Maclaurin series of the complete elliptic integral of the first kind (in the following, the argument of $K$ is the elliptic modulus) $$ K(x)=\frac{\pi}{2}\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2 x^n \tag{2}$$ such that the LHS of $(1)$ blatantly is $\frac{2}{\pi}\int_{0}^{1}K(x^2)\,dx$ or $$ \frac{1}{\pi}\int_{0}^{1}\frac{K(x)}{\sqrt{x}}\,dx.\tag{3}$$ Due to the generating function for Legendre polynomials, both $K(x)$ and $\frac{1}{\sqrt{x}}$ have very simple FL (Fourier-Legendre) expansions, namely $$ K(x)=\sum_{m\geq 0}\frac{2}{2m+1}P_m(2x-1),\qquad \frac{1}{\sqrt{x}}=\sum_{m\geq 0}2(-1)^m P_m(2x-1) \tag{4} $$ hence by the orthogonality relation $\int_{0}^{1}P_n(2x-1)P_m(2x-1)\,dx=\frac{\delta(m,n)}{2n+1}$ we get $$ \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{1}{2n+1} = \frac{4}{\pi}\sum_{m\geq 0}\frac{(-1)^m}{(2m+1)^2}=\frac{4G}{\pi}\tag{5}$$ QED.

This approach is powerful enough to let you compute much worse.

For some integrals: $$\color{blue}{\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^2}{1+x^2}\right)\frac{dx}{x}=\pi G}$$ $$\color{red}{\int_0^\frac{\pi}{2} x\ln\left(\cot\left(\frac{x}{2}\right)\left(\frac{\sec x}{2}\right)^4\right)dx=\pi G}$$