First, I had put $\textstyle{\displaystyle{\sum_{n=1}^{\infty}{\lim_{u\rightarrow\infty}{\frac{(-1)^nu^{n+s}}{(n-1)!(n+s)}}}}}$ into wolfram alpha and got nothing.
Then, I thought about inter-changing the limit and the summation which will give us
$\textstyle{\displaystyle{\lim_{u\rightarrow\infty}{\sum_{n=1}^{\infty}\frac{(-1)^nu^{n+s}}{(n-1)!(n+s)}}}}$
which did gave me a result again from wolfram alpha.
Putting the sum in wolfram alpha gave me
$\begin{align}&-\Gamma(s+1,0,u)\\
&=-\Gamma(s+1,0)+\Gamma(s+1,u)\\
&=\Gamma(s+1,u)-\Gamma(s+1)\end{align}$
Then taking the limit gives us
$\textstyle{\displaystyle{\lim_{u\rightarrow\infty}(\Gamma(s+1,u)-\Gamma(s+1))}}$
$=-\Gamma(s+1)$
However, I am not sure that I can really just interchange the limit and summation. While searching about this I got this post on this site, which says
$\begin{align}\textstyle\displaystyle{\lim_{n\rightarrow\infty}\sum_{m=1}^{\infty}f(m,n)\geq\sum_{m=1}^{\infty}\lim_{n\rightarrow\infty}f(m,n)}\end{align}$
Applying this inequality on my sum gives us
$\textstyle\displaystyle{\sum_{n=1}^{\infty}\lim_{u\rightarrow\infty}{\frac{(-1)^nu^{n+s}}{(n-1)!(n+s)}}\leq -\Gamma(s+1)}$
However, this doesn't really give the value of the sum. So, my question is
What is the value of $\textstyle{\displaystyle{\sum_{n=1}^{\infty}{\lim_{u\rightarrow\infty}{\frac{(-1)^nu^{n+s}}{(n-1)!(n+s)}}}}}$?
Best Answer
In general the growth of $u$ is not related to the growth of $n$ (the summation argument). The correct answer to this question is that the $\sum_{n=1}^{\infty}{\lim_{u\rightarrow\infty}{\frac{(-1)^nu^{n+s}}{(n-1)!(n+s)}}}$ has no definite value.
In contrast, $$\lim_{u\rightarrow\infty}\sum_{n=1}^{\infty}{{\frac{(-1)^nu^{n+s}}{(n-1)!(n+s)}}}$$ exists as the summation parameter vanishing.
If we denote: $$f(u,n)=\frac{(-1)^nu^{n+s}}{(n-1)!(n+s)}$$ then $$g(u)=\sum_{n=1}^{\infty}f(u,n)$$
$$g'(u)=f(u,n)_u'=\frac{(-1)^nu^{n+s-1}}{(n-1)!}$$ $$\sum_{n=1}^{\infty}f(u,n)_u'=-e^{s-u}$$ So after integration: $$\sum_{n=1}^{\infty}f(u,n)=e^{s-u} + C(s)$$ Now $$\lim_{u\rightarrow\infty}\sum_{n=1}^{\infty}f(u,n)=\lim_{u\rightarrow\infty}e^{s-u}+C(s)=C(s)$$
$C(s)$ here is $-\Gamma(s+1,0)$. Incomplete $\Gamma$ function https://en.wikipedia.org/wiki/Incomplete_gamma_function.