Compute $\sum_{m>n=1}^{\infty} \frac{1}{m!n!}$ Compute the series
$$\sum_{m>n=1}^{\infty} \frac{1}{m!n!}$$

We have $$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{  m!\,n!} = \left(\frac{1}{1!} + \frac{1}{2!}+ \frac{1}{3!}+\cdots\right)\left(\frac{1}{1!} + \frac{1}{2!}+ \frac{1}{3!}+\cdots\right)=(e-1)^2$$
We must substract from that the "diagonal" terms: $$\sum_{n=1}^{\infty} \left(\frac{1}{n!}\right)^2 = I_0(2) - 1 \approx 2.2795853-1$$ (ref ; $I_0(\cdot)$  is the modified Bessel function of the first kind) and divide by two to get the desired region. Hence the result is
$$\sum_{n=1}^{\infty}\sum_{m=n+1}^{\infty} \frac{1}{  m!\,n!}=\frac{(e-1)^2 - (I_0(2)-1)}{2}= \frac{e^2-I_0}{2}+1-e \approx 0.83645357$$
