Consider the sequence $a\_n=2^{2n}\binom{2n}n^{-1}$. [Stirling's approximation](https://en.wikipedia.org/wiki/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?
>  $$\sum\_{n=0}^{\infty}\left(\frac{\pi}{2a\_n}-\frac{a\_n}{2n+1}\right)=1.$$
> 
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> 



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](https://en.wikipedia.org/wiki/Abel's_theorem).


