4
$\begingroup$

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 $\gamma$?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – J. M. is a poor mathematician May 16 '16 at 11:40
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ "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. $\endgroup$ – J. M. is a poor mathematician May 16 '16 at 12:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.