Are these new series formulae for $\zeta(2)$?

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.

• Mathematica knows both. – eyeballfrog Aug 13 '19 at 17:08
• We have: $$\sum_{n=1}^\infty \frac{x^n}{n^2 \binom{2n}{n}}=2\arcsin^2 \frac{\sqrt{x}}{2}$$ You can check your particular cases using this general formula – Yuriy S Aug 16 '19 at 11:35
• To answer the question, no, these results are not new, since the general Taylor series for $\arcsin^2 y$ has been known for a long time – Yuriy S Aug 16 '19 at 11:38
• @YuriyS Thank you for the comment! If you could elaborate on that in the form of an answer I will be happy to accept it – Klangen Aug 16 '19 at 11:42
• As a (rather silly) instance of this formula, one may write $$\zeta{2} = \frac{25}{12}\sum_{n=1}^\infty \frac{1}{n^2 \binom{2n}{n}} \left(\frac{{5}-\sqrt{5}}{2}\right)^n.$$ – Semiclassical Aug 16 '19 at 20:49

$$\displaystyle \sum_{n=1}^\infty \frac{x^n}{n^2 \binom{2n}{n}}=2\left[\sin^{-1} \frac{\sqrt{x}}{2} \right]^2$$
Just note that in the result given in answer, you need to replace $$x$$ by $$\frac{\sqrt{x}}{2}$$ and also make sure that $$0 \leq x \leq 4$$