$\lim_{n\rightarrow \infty} \frac{1^4+2^4+...+n^4}{n^5}=\frac{1}{5}$ using integral(lower-/upper- or Riemann-sums) how can I show that $$
\lim_{n\rightarrow \infty} \frac{1^4+2^4+...+n^4}{n^5}=\frac{1}{5}$$
using integral(lower-/upper- or Riemann-sums).
I tried the following:
$\lim_{n\rightarrow \infty} \frac{1^4+2^4+...+n^4}{n^5}= \lim_{N\rightarrow\infty}\frac{1}{N^5} \lim_{N \rightarrow\infty} \sum_{n=1}^Nn^4=\frac{\frac{N^5}{5}}{N^5}=\frac{1}{5}$
But it feels like I am missing some steps here, especially why $\lim_{N \rightarrow \infty}\sum_{n=1}^N n^4=\frac{N^5}{5}$
How can I improve my try?
Thank you in advance!

Hint
$$\frac{\sum_{k=1}^n k^4}{n^5}=\frac{1}{n} \sum_{k=1}^n \left(\frac{k}{n} \right)^4=\frac{1}{n} \sum_{k=1}^n f\left(\frac{k}{n} \right)$$
with $f(x)=x^4$, so you can recognize a Riemann sum.
