ca.classical-analysis-and-odesreal-analysisreference-requestsequences-and-series
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?
$$\sum_{n=0}^{\infty}\left(\frac{\pi}{2a_n}-\frac{a_n}{2n+1}\right)=1.$$
I think that this is "amazing claim" number 1 in van der Poorten's paper "A proof that Euler missed ...."
The sum is equal to $$ \int\limits_{0}^{\infty}\frac{\exp(-x)}{1+W_0(x)}\,\mathrm{d}x = 0.7041699604... $$ where $W_0(x)$ is the Lambert-$W$ function.
Reference
Stephen Finch. "Errata and Addenda to Mathematical Constants [Math.HO]", §6.11. Iterated Exponential Constants, pg. 66.
Two interesting divergent sums from this reference are as well $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}e^{-x}\log(x/|W_{0}(i x)|)\mathrm{d}x = 0.3233674316...$$ which seems to be also the cosine and sine integrals of the derivatives $W'_{0}(x) = \frac {W_{0}(x)}{x(1 + W_{0}(x))}$ and $-W''_{0}(x)$ respectively $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}\mathrm{cos}(x) W'_{0}(x)\mathrm{d}x = -\int\limits_{0}^{\infty}\mathrm{sin}(x) W''_{0}(x)\mathrm{d}x$$ as it can be seen in the following Pari GP v2.14.0 lines
although, as it is pointed out, a rigorous proof is not yet known.
And this variation, $$\sum_{n=1}^\infty (-1)^{n-1} (2n-1)^{2n} = \frac{1}{2i}\int\limits_{0}^{\infty}e^{-x}(g(ix)-g(-ix))\mathrm{d}x = 0.0111203007...$$ where $g(x) = W_{0}(x)/(1 + W_{0}(x))^3$
Best Answer
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.