Calculate the following sequence: $\lim_{n\to \infty} a_n = \frac{1}{n^5}(1^4 +2^4+....+n^4)$.
I know how to solve this using Riemann sums of the function $x^4$.
I was wondering if there's another way using other methods. Any help would be appreciated.
HINT
By Stolz-Cesaro
$$\lim_{n\to \infty} \frac{b_n}{c_n}=\lim_{n\to \infty} \frac{1^4 +2^4+…+n^4}{n^5}=\lim_{n\to \infty} \frac{b_{n+1}-b_n}{c_{n+1}-c_n}=\lim_{n\to \infty} \frac{(n+1)^4}{(n+1)^5-n^5 }$$
and recall that
$$(n+1)^5-n^5=5n^4+10n^3+10n^2+5n+1$$
There are constants $c_0,c_1,c_2,c_3,c_4$ such that $$n^4=\sum_{r=0}^4c_r\binom{n}r\tag1$$
That leads to:$$n^{-5}\sum_{k=1}^nk^4=n^{-5}\sum_{k=1}^n\sum_{r=0}^4c_r\binom{k}r=n^{-5}\sum_{r=0}^4c_r\sum_{k=1}^n\binom{k}r=n^{-5}\sum_{r=0}^4c_r\binom{n+1}{r+1}$$
This on base of the equality:$$\sum_{k=r}^{n}\binom{k}{r}=\binom{n+1}{r+1}$$
Taking the limit on both sides gives $\frac{c_4}{5!}$ as outcome, so it remains to find $c_4$ on base of $(1)$ which is not difficult:$$c_4=4!=24\text{ and consequently: }\lim_{n\to\infty}n^{-5}\sum_{k=1}^nk^4=\frac{4!}{5!}=\frac15$$
Actually more generally we find by this method:$$\lim_{n\to\infty}n^{-m-1}\sum_{k=1}^nk^m=\frac{m!}{(m+1)!}=\frac1{m+1}$$
The Binomial Theorem gives $$ (k+1)^5-k^5=5k^4+10k^3+10k^2+5k+1\tag1 $$ Summing both sides yields $$ \begin{align} (n+1)^5-1^5 &=\sum_{k=1}^n\left(5k^4+10k^3+10k^2+5k+1\right)\\ &=5\sum_{k=1}^nk^4+O\!\left(n^4\right)\tag2 \end{align} $$ since $$ \begin{align} \sum_{k=1}^n\left(10k^3+10k^2+5k+1\right) &\le\sum_{k=1}^n\left(10n^3+10n^2+5n+1\right)\\ &=10n^4+10n^3+5n^2+n\\[6pt] &=O\!\left(n^4\right)\tag3 \end{align} $$ Dividing $(2)$ by $5n^5$ and taking the limit as $n\to\infty$ leads to $$ \frac15=\lim_{n\to\infty}\frac1{n^5}\sum_{k=1}^nk^4\tag4 $$
