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? $$\sum_{n=0}^{\infty}\left(\frac{\pi}{2a_n}-\frac{a_n}{2n+1}\right)=1.$$