Sum to Infinity of 2x^2n over n by 2n Choose n

Theorem
For $\cmod x < 1$:


 * $\ds \frac {2 x \arcsin x} {\sqrt {1 - x^2} } = \sum_{n \mathop = 1}^\infty \frac {\paren {2 x}^{2 n} } {n \dbinom {2 n} n}$

Proof
By Gregory Series:
 * $\ds \arctan t = \sum_{m \mathop = 0}^\infty \frac {\paren {-1}^m t^{2 m + 1} } {2 m + 1}$

Let $t = \dfrac x {\sqrt {1 - x^2} }$.

Let $y = \arcsin x$.

Then:

Hence $\arctan t = \arcsin x$.

We have:

It remains to show the the coefficient of $x^{2 n}$ on the is equal to $\dfrac {2^{2 n} } {n \dbinom {2 n} n}$, that is:
 * $\ds 2 \sum_{r \mathop = 1}^n \frac {\paren {-1}^{r - 1} } {2 r - 1} \dbinom {r + n - r - 1} {r - 1} = \frac {2^{2 n} } {n \dbinom {2 n} n}$

The above is generated by picking, for each $m > 0$, the corresponding $k = n - m$ from the right sum $\ds \sum_{k \mathop = 0}^\infty \dbinom {m + k - 1} {m - 1} x^{2 k}$.

We have:

Hence the result.