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There is apparently a curious connection between Euler-Mascheroni constant $\gamma$ and $e$ in the form of an infinite series and continued fraction:
$$e \gamma=e \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!~n}-\cfrac{1}{2-\cfrac{1}{4-\cfrac{4}{6-\cfrac{9}{8-\cfrac{16}{10-\cdots}}}}}$$
As can be seen, the partial denominators and numerators have the form $2n$ and $-n^2$ respectively.
How can we prove this? Might it be a useful method to compute $\gamma$?
Never mind, thanks to the comment by J. M. I found the source of this expression.
The series are connected to the exponential integral:
$$\text{Ei}(t)=-\int_{-t}^{\infty} \frac{e^{-p}}{p} dp=\gamma+\log |t|+\sum_{n=1}^{\infty} \frac{t^n}{n!n}$$
The continued fraction turns out to be a particular case of incomplete Gamma function:
$$\Gamma (0,t)=\int_t^{\infty} \frac{e^{-p}}{p} dp=\cfrac{\exp(-t)}{t+1-\cfrac{1}{t+3-\cfrac{4}{t+5-\cfrac{9}{t+7-\cdots}}}}$$
So, we obtain:
$$\text{Ei}(-1)+\Gamma (0,1)=0$$
$$\gamma-\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!n}+\cfrac{\exp(-1)}{2-\cfrac{1}{4-\cfrac{4}{6-\cfrac{9}{8-\cdots}}}}=0$$
Is still don't know how the first two expressions for $\text{Ei}(t)$ and $\Gamma (0,t)$ are proven, but I'll try to find the references.
