TITLE: Two divergent series conspiring?
QUESTION [11 upvotes]: Consider the sequence $a_n=2^{2n}\binom{2n}n^{-1}$. Stirling's approximation shows that $a_n\sim \sqrt{\pi n}$, thus
$$\sum_{n\geq0}\frac{\pi}{2a_n}\qquad \text{and} \qquad
\sum_{n\geq0}\frac{a_n}{2n+1}$$
are both divergent series. However, their difference should converge with terms of order $\sim\frac1{n^{3/2}}$.

Question. In fact, is this true?
  $$\sum_{n=0}^{\infty}\left(\frac{\pi}{2a_n}-\frac{a_n}{2n+1}\right)=1.$$

REPLY [29 votes]: 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.