How to prove this continued fraction connection between $\gamma$ and $e$?

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 $\gamma$?

• I note that the series in your expression is in fact related to the so-called exponential integral, which indeed features the Euler-Mascheroni constant in one of its definitions. – J. M. is a poor mathematician May 16 '16 at 11:40

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.

• "I still don't know how the first two expressions for $\mathrm{Ei}(t)$ ... are proven" - one possible route is to use a three-term recurrence relation for the Kummer confluent hypergeometric function; this can be transformed into an equivalent continued fraction identity, and the incomplete gamma function/exponential integral then drops out as a special case. – J. M. is a poor mathematician May 16 '16 at 12:58