Let $\zeta(n)$ denote the Riemann zeta function defined for positive integers greater than $1$ by its usual infinite series. Thus, $\zeta(2)=\sum_{k=1}^\infty\frac{1}{k^2}$. Many formulae exist involving $\zeta(2)$, including the Apéry-like fast-converging series: $$ \zeta (2)=3\sum _{{n=1}}^{{\infty }}{\frac {1}{n^{{2}}{\binom {2n}{n}}}}. $$ Recently I have found the following similar-looking series:
$$ \zeta (2)=\frac83\sum _{{n=1}}^{{\infty }}{\frac {2^{n-1}}{n^{{2}}{\binom {2n}{n}}}}, $$ $$ \zeta (2)=\frac94\sum _{{n=1}}^{{\infty }}{\frac {3^{n-1}}{n^{{2}}{\binom {2n}{n}}}} $$ and $$ \zeta (2)=\frac43\sum _{{n=1}}^{{\infty }}{\frac {4^{n-1}}{n^{{2}}{\binom {2n}{n}}}}. $$
Are these series already known? A quick internet search yields no such results.
EDIT forgot to add the second series.
Please see the following answer for the proof of the fact
$\displaystyle \sum_{n=1}^\infty \frac{x^n}{n^2 \binom{2n}{n}}=2\left[\sin^{-1} \frac{\sqrt{x}}{2} \right]^2$
Just note that in the result given in answer, you need to replace $x$ by $\frac{\sqrt{x}}{2}$ and also make sure that $0 \leq x \leq 4$