<s>[INST] Calculate the following limit: $\lim_{n\to \infty} a_n = \frac{1}{n^5}(1^4 +2^4+....+n^4)$ 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.
 [/INST] 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
$$
\frac15=\lim_{n\to\infty}\frac1{n^5}\sum_{k=1}^nk^4\tag4
$$
</s>