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

  • $\begingroup$ Would it be correct to assume that you have some numerical evidence that suggests this is the case? (otherwise this seems wildly improbable). $\endgroup$ – Anthony Quas Mar 4 '17 at 1:27
  • $\begingroup$ Absolutely, yes. $\endgroup$ – T. Amdeberhan Mar 4 '17 at 1:28
  • $\begingroup$ Experiment makes it clear that the terms in the sum are $\sqrt{\pi/n}(n^{-1}/8-3n^{-2}/32+O(n^{-3}))$. $\endgroup$ – Neil Strickland Mar 4 '17 at 2:14

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.

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    $\begingroup$ Richard: that is cool and clean. $\endgroup$ – T. Amdeberhan Mar 4 '17 at 5:05

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.

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    $\begingroup$ This analysis is almost what I commented on in the question, although not in detail. Anyhow, the convergence was clear. The main question is to show the sum actually equal to $1$. $\endgroup$ – T. Amdeberhan Mar 4 '17 at 3:44
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    $\begingroup$ Sorry, I was too tired when I read your question and missed that the main point was the evaluation of the sum. I was also mislead by you saying "should converge" instead of saying "converges". I am glad Richard Stanley has clarified the situation, and I agree his proof is cool and clean. $\endgroup$ – GH from MO Mar 4 '17 at 9:25
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    $\begingroup$ Of course, I thank you for the details in your answer. $\endgroup$ – T. Amdeberhan Mar 4 '17 at 13:33

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