On the series $\sum_{n=0}^{\infty}1/(n!)^x$ I was playing with functions defined as infinite sums, when i came across this function:$$f(x)=\sum_{n=0}^{\infty}\frac{1}{(n!)^x}$$
this series converges absolutely for $x>0$ (You can use the ratio test to check that.), here is some values of $f(x)$ (i used a python script to calculate them):
$$\begin{array} {|r|r|}\hline f(1) & e=2.71828.. \\ \hline f(2) &  2.2795..\\ \hline f(10)  &  2.00097\\ \hline f(50)&2.000...1\\ \hline  \end{array}$$
You may notice that as $x$ gets bigger $f(x)$ approaches $2$. I have a proof of that claim, but i’m not sure if it is true, here is my proof:
$$f(x)=\sum_{n=0}^{\infty}\frac{1}{(n!)^x}=\frac{1}{(0!)^x}+ \frac{1}{(1!)^x}+ \frac{1}{(2!)^x}+... $$
$$f(x)=1+1+ \frac{1}{(2!)^x} +...$$
For $n>1$ :
$$\lim_{x\to\infty}\frac{1}{(n!)^x}=0$$
So :$$ \lim_{x\to\infty}f(x)=1+1=2$$
 A: You cannot conclude directly that since the limit of each term with $n\geq 2$ is $0$, that the infinite series of all the terms $n\geq 2$ also goes to $0$.  This needs a separate argument.  Here is one such argument:
First note that when $x \geq 1$, we have $$0 \leq \frac{1}{(n!)^x} \leq \frac{1}{n!}$$ and therefore, for any $x \geq 1$ and $N \geq 2$, we have
$$0 \leq \sum_{n=N+1}^\infty \frac{1}{(n!)^x} \leq \sum_{n=N+1}^\infty \frac{1}{n!},$$
which will tend to $0$ as $N \to \infty$... and that's pretty much all we need to set up the argument.

Specifically, we have for any $x \geq 1$ and $N \geq 2$, $$0 \leq f(x) - 2 = \sum_{n=2}^\infty \frac{1}{(n!)^x} = \sum_{n=2}^N\frac{1}{(n!)^x} + \sum_{n=N+1}^\infty \frac{1}{(n!)^x} \leq \sum_{n=2}^N\frac{1}{(n!)^x} + \sum_{n=N+1}^\infty \frac{1}{n!}$$
By applying $\lim_{N\to\infty}\limsup_{x\to\infty}$ to this compound inequality, we have $$0 \leq \limsup_{x\to\infty} (f(x)-2) \leq \lim_{N\to\infty}\limsup_{x\to\infty}\left(\sum_{n=2}^N\frac{1}{(n!)^x} + \sum_{n=N+1}^\infty \frac{1}{n!}\right) = \lim_{N\to\infty}\sum_{n=N+1}^\infty \frac{1}{n!} = 0$$
from which it follows that $$0 \leq \liminf_{x\to\infty} (f(x)-2) \leq \limsup_{x\to\infty} (f(x)-2) = 0$$
so the limit exists and $\lim_{x\to\infty} f(x) = 2$.

The above is, in principle, the same method given by the Dominated Convergence theorem, but like so many other integration theorems, they become much easier to prove in the limited context of sums and counting measures.
