I have the infinite sum:
\begin{align}
S_n = \frac{n^5!}{(n+1)^5!}+\frac{n^5!}{(n+2)^5!}+\frac{n^5!}{(n+3)^5!}+\cdots
\end{align}
I must find a upper bound for $S_n$, that is, $0<S_n<f(n)$, where $f(n)$ is a function that $\lim_{n\to\infty} f(n) = 0$, because, I'm not sure if this is allowed in a infite series, but $n^5!/(n+1)^5!$ goes to zero as $n$ tends to infinite, and the same happens for all other terms, so $\lim_{n\to \infty}S_n = 0 \:\:(?)$. So exists a $f(n)$ such that, $0<S_n<f(n)$ where $f(n)\to 0 $ as $n\to \infty$
For example: try to estimate (find a upper bound) the infinite series: $R_n = n!/(n+1)! + n!/(n+2)!+\cdots$
I can cancel the terms after expanding the factorial, resulting in $R_n = 1/(n+1)+1/(n+2)(n+1)+\cdots<1/(n+1)+1/(n+1)(n+1)+\cdots = 1/n$ (because it's a geometric series). So we have that $0<R_n<1/n$. And that's enough. Now I must do the same for $S_n$ defined above, but I can't cancel terms after expanding the factorial.
Best Answer
It holds that $$ \frac{n^5!}{(n+k)^5!}\le\frac{1}{(n^5+k)(n^5+k+1)}=\frac{1}{n^5+k}-\frac{1}{n^5+k+1} $$ since $(n+k)^5\ge n+k+1$, and hence $$ S_n = \sum_{k=1}^\infty \frac{n^5!}{(n+k)^5!}\le \sum_{k=1}^\infty \left(\frac{1}{n^5+k}-\frac{1}{n^5+k+1}\right)=\frac{1}{n^5+1}\to 0. $$