Algorithm used by Mathematica for evaluating partial sums

Question

Today, while using mathematica, I found that: $$\sum_{x\geq n\geq0}\frac{1}{n!}=\frac{e\Gamma(x+1,1)}{\Gamma(x+1)}$$ Where $\Gamma(x,y)$ is the Incomplete Gamma function.
It is well known that $$\sum_{x\geq0}\frac{1}{n!}=e$$ But which algorithm(or formula) did Mathematica use? If the algorithm is too complex for human use, how can we prove the first equality?
Another result it gave was $$\sum_{m\geq n\geq 1}\frac{1}{n\,n!}=\frac{(\mathrm{Ei}(1)-\gamma)m(m+1)!+(\mathrm{Ei}(1)-\gamma)(m+1)!-_2F_2(1,m+1;m+2,m+2;1)}{(m+1)(m+1)!}$$ Where $\mathrm{Ei}$ is the Exponential integral and $F$ is the Generalized hypergeometric function. This looks very hard to evaluate by hand, is there any formula or algorithm to evaluate this? I shouldn't say this on MSE, but I don't know how to do this, I can't do even a step.

Answer 1

In any manner, the expression can rewrite $$\sum_{m\geq n\geq 1}\frac{1}{n\,n!}=-m \Gamma (m) \, _2\tilde{F}_2(1,m+1;m+2,m+2;1)+\text{Ei}(1)-\gamma$$ where appears the regularized generalized hypergeometric function.

This is not easier at all to compute but, at least, the formula is shorter.

If you want a shortcut calculation, considering $$f_m=\sum_{m\geq n\geq 1}\frac{1}{n\,n!}\qquad \text{and} \qquad g_m=\sum_{m\geq n\geq 1}\frac{1}{,n!}$$ you could notice that the ratio $$h_m=\frac { f_m}{g_m} \sim k\, $$

Computed for $1 \leq m \leq 20$, a quick and dirty linear regression gives, with $R^2= 0.999956$, $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ k & 0.767238 & 0.001200 & \{0.764718,0.769758\} \\ \end{array}$$

The asymptotic value of $k$ is $$0.766988354079434252937356586124784190417745778500724543367683970$$