Is $2\pi\sum_{n=0}^\infty\frac1{n!(n+1)!}$ equal to $5/\pi$?

Question

I was working through some contour integration questions, and when finding the residues of the integral $\int e^{z+1/z}\,dx$, I found that it was equal to the infinite sum $$2\pi\sum_{n=0}^\infty\frac1{n!(n+1)!}$$ When I put the integral into a calculator, it came out as a number so close to 10 that I assumed it was an error due to the calculator using a sum, but then when I put the sum into the calculator it was different. Is this an error, or is the sum actually equal to $\frac5\pi$?

Answer 1

This is just a coincidence. The infinite sum is a Bessel function $I_1(2)=1.590636\dots$, and this times $2\pi$ is $$9.994266\dots$$

Answer 2

Not too much of a coincidence. By the integral representation for modified Bessel functions of the first kind,

$$ 2\pi\sum_{n\geq 0}\frac{1}{n!(n+1)!}=\int_{-\pi}^{\pi}e^{2\cos\theta}\cos(\theta)\,d\theta $$ is bounded between $$ \int_{-\pi/2}^{\pi/2}e^2 \cos^3(\theta)\,d\theta=\frac{4}{3}e^2\quad\text{and}\quad \int_{\mathbb{R}}e^{2-3\theta^2/2}\,d\theta=\sqrt{\frac{2\pi}{3}}e^2 $$ both of which are pretty close to $10$.