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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 $\textstyle{\displaystyle{\sum_{n=1}^{\infty}{\lim_{u\rightarrow\infty}{\frac{(-1)^nu^{n+s}}{(n-1)!(n+s)}}}}}$?

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  • $\begingroup$ When $n=1$, the limit diverges. So wouldn't the expression be undefined? $\endgroup$
    – Alan Abraham
    Commented Aug 7, 2021 at 4:39
  • $\begingroup$ @Alan Abraham. The limit diverged at all $n$ values. That is the problem. If you apply the limits individually, you get, $-\infty+\infty-\infty+\cdots$. And remember $\infty-\infty$ is indeterminate. $\endgroup$
    – Rounak Sarkar
    Commented Aug 7, 2021 at 5:07
  • $\begingroup$ That works if the summation is inside the limit. I don't think it does the other way around, however. $\endgroup$
    – Alan Abraham
    Commented Aug 7, 2021 at 5:16
  • $\begingroup$ What do you mean? $\sum_{n=1}^{\infty}{\lim_{u\rightarrow\infty}{\frac{(-1)^nu^{n+s}}{(n-1)!(n+s)}}}=\lim_{u\rightarrow\infty}\frac{(-1)^1u^{1+s}}{(1-1)!(1+s)}+\lim_{u\rightarrow\infty}\frac{(-1)^2u^{2+s}}{(2-1)!(2+s)}+\cdots=-\infty+\infty-\cdots$ $\endgroup$
    – Rounak Sarkar
    Commented Aug 7, 2021 at 5:49
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    $\begingroup$ If each limit doesn't exist, it certainly doesn't make sense to sum them.... $\endgroup$
    – Alan
    Commented Aug 7, 2021 at 6:35

1 Answer 1

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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.

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  • $\begingroup$ I don't see why $\lim_{u\to\infty} e^{s-u}=C(s)$. Shouldn't it equal $e^s\lim_{u\to\infty} e^{-u}=0$? $\endgroup$
    – Alan Abraham
    Commented Aug 7, 2021 at 13:28
  • $\begingroup$ Just added correction to answer. $\endgroup$
    – Gevorg Hmayakyan
    Commented Aug 7, 2021 at 17:50

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