How to find $\mathrm{\sum_{n=1}^\infty E_{n}(n)= \lim_{k\to\infty}\int_1^\infty \frac{e^{kt+t}t^k-e^t}{e^{kt+2t}t^{k+1}-e^{kt+t}t^k}dt=.26929…}$?

As seen from this

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Equating the arguments of gamma type functions and inducing a transformation onto them often gives nice results. These results are usually hard to evaluate using non-trivial methods. Here is a generalized exponential integral function version. Here is the series for it, although it was said to diverge for the seemingly convergent series. Incomplete gamma function information:

$$\mathrm{S_E=\sum_{n=1}^\infty E_{n}(n)=\sum_{n=1}^\infty n^{n-1}Γ(1-n,n)=\sum_{n=1}^\infty n^{n-1} \int_n^\infty t^{-n}e^{-t}dt=\sum_{n=1}^\infty \ \int_1^\infty \frac{dt}{e^{nt}t^n} =\quad\lim_{k\to\infty}\int_1^\infty \frac{e^{kt+t}t^k-e^t}{e^{kt+2t}t^{k+1}-e^{kt+t}t^k}dt= 0.2692923…}$$

I will work on this more, but wonder if there is an integral representation or a way to evaluate the sum using any method or at least get rid of the limit. Here is proof of the partial sums. Please do not just expand the summation and say that is the answer or some other related “trivial” evaluation, as I am looking for an “actual” answer. It can use any well know functions and does not need to be in closed form, but simplify as much as possible. How can I write S without using a sum? Please correct me and give me feedback!

Solution 1:

$$E_n(n)=\int_1^{\infty } e^{-n t} t^{-n} \, dt\tag{1}$$

$$\sum _{n=1}^{\infty } e^{-n t} t^{-n}=\frac{1}{t\ e^t-1}\tag{2}$$

$$S=\sum\limits_{n=1}^\infty E_{n}(n)=\int\limits_1^\infty\frac{1}{t\ e^t-1}\,dt=0.269292\tag{3}$$

The result in (3) above was derived using numerical integration via the Mathematica NIntegrate function. Also see Wolfram Alpha evaluation of the integral.