Two divergent series conspiring?

We have $$ f(x):=\sum_{n\geq 0}\frac{x^{2n}}{a_n} = \frac{1}{\sqrt{1-x^2}} $$ and $$ g(x):=\sum_{n\geq 0} \frac{a_n}{2n+1}x^{2n} = \frac{\sin^{-1}x} {x\sqrt{1-x^2}}. $$ It is routine to compute that $$ \lim_{x\to 1-}\left(\frac 12\pi f(x)-g(x)\right)=1 $$ and then apply Abel's theorem.

Yes, the difference of the two series converges absolutely. First, note that the refined Stirling approximation $$ n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n(1+O(n^{-1}))$$ yields $a_n=\sqrt{\pi n}(1+O(n^{-1}))$, hence also $a_n^2=\pi n(1+O(n^{-1}))$. Therefore, $$ \left|\frac{\pi}{2a_n}-\frac{a_n}{2n+1}\right| = \frac{\bigl|\pi(2n+1)-2\pi n(1+O(n^{-1}))\bigr|}{(4n+2)a_n}=\frac{O(1)}{(4n+2)a_n}=O(n^{-3/2}),$$ and the claim follows by the convergence of $\sum_{n=1}^\infty n^{-3/2}$.

Added. I missed that the main point of the question was the evaluation of the difference series. For this, see Richard Stanley's response.