Say I wanted to simplify
$$ \sum\limits_{n=1}^{\infty} \frac{n!}{1000^n n^{1000}} $$
Now I could cancel one n, but the real question is, will $1000^n$ show up as a factor in the factorial of $n!$ because n goes to infinity?
Now you would think it would not, because if you took an integer like 10, you know that $1000^{10}$ is not a factor of $1000!$
Does that change for $n=\infty$ ?
Thinking about it for a little bit more I concluded some basic observations:
As Sophie said: $$ \lim_{n \rightarrow \infty} \frac{n!}{1000^n n^{1000}} = \infty $$ but $$ \frac{n! \frac{1}{1000^n}}{n^{1000}} < \frac{n!}{n^{1000}} $$
and$$ \lim_{n \rightarrow \infty} \frac{n!}{n^{1000}} = \infty $$
So I believe the statement if the sum converges or not does not give us more information about if that factor is going to be canceled out.
Not knowing if intuitive thinking is appropriate in this context, I'd say it boils down to the question, if $ \infty !$ contains infinity as a factor ( since $1000^{\infty} = \infty $) , but again, I do not think you can treat infinity like that.
$\displaystyle \lim_{n\to\infty}\left(\frac{n!}{1000^nn^{1000}}\right)=\infty$
So the sum diverges.