Can you solve this two limits (with demostration, obviously)?
Ei(n) is the exponential integral function
I'd like to suggest a start but not sure it's sound. Perhaps others can comment.
First, clearly define the integral:
$\displaystyle \text{Ei}_n(x)=\int_1^{\infty}\frac{e^{xt}}{t^n}dt$
so need to consider:
$\displaystyle \sum_{n=1}^{\infty}\int_1^{\infty}\frac{e^{xt}}{t^ n}dt-\sum_{n=1}^{\infty}\frac{e^{-x}}{n+x}$
Now express the last expression as the same type integral:
$\displaystyle \frac{e^{-x}}{n+x}=\int_1^{\infty}e^{1-t}\frac{e^{-x}}{n+x}dt$
Combining both sums, I obtain:
$\displaystyle \sum_{n=0}^{\infty}\int_1^{\infty}\left( \frac{e^{-xt}}{t^n}-\frac{e^{-x}e^{1-t}}{n+x}\right)dt$
Now, at this point, how about using the Integral Test? I'm however not sure that's valid and I'd need to work on that part. If it were valid, I could then write:
$\displaystyle \int_{1}^{\infty}\int_1^{\infty} \frac{e^{-xt}}{t^n}-\frac{e^{-x}e^{1-t}}{n+x}dtdn$
Switching the order:
$\displaystyle \int_{1}^{\infty}\int_1^{\infty} \frac{e^{-xt}}{t^n}-\frac{e^{-x}e^{1-t}}{n+x}dndt$
I get:
$\displaystyle \int_{1}^{\infty}\left\{\left(-\frac{1}{\ln(t)}e^{-n\ln(t)}\Biggr|_1^{\infty}\right)-e^{-x}e^{1-t}\ln(n+x)\Biggr|_1^{\infty}\right\}dt$
and that last term diverges. So if the steps I've taken so far are valid, and I could apply the integral test in this fashion, then I would conclude the original sum diverges.
. . . Gee whiz, you guys are way better at this than me but I'm big at tries.