Sum Of Fourth Powers?

Relevant wiki: Riemann Sums

See Reimann Sums $\mathcal U=\displaystyle\lim_{n\to\infty}\dfrac{n^{p+1}}{\underbrace{1^p+2^p+3^p\cdot +n^p}_{\displaystyle\sum_{r=1}^{n}r^{p}}}$

$~~~=\dfrac{1}{\displaystyle\lim_{n\to\infty}\dfrac 1n\displaystyle\sum_{r=1}^{n}\left(\dfrac rn\right)^{p}}=\dfrac{1}{\displaystyle\int_0^1 x^p\mathrm{d}x}=\boxed{p+1}$

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For this question ,put $p=4$ , so that:-

$\lim_{n\to\infty} \dfrac{n^5}{1^4+2^4+\cdots+ n^4} =\boxed{\color{#D61F06}{5}}$

We note that $\begin{aligned} \lim_{n \to \infty} \dfrac {n^{m+1}}{1^m + 2^m + 3^m + ... + n^m} = \frac {a_n}{b_n} \end{aligned}$ is such that $0 < b_1 < b_2 < ... < b_n < ...$ and $\lim_{n \to \infty} b_n = \infty$ is a $\infty/\infty$ case and we can apply Stolz–Cesàro theorem as follows:

$\begin{aligned} L & = \lim_{n \to \infty} \dfrac {n^{m+1}}{1^m + 2^m + 3^m + ... + n^m} \\ & = \lim_{n \to \infty} \frac {a_{n+1}-a_n}{b_{n+1}-b_n} \\ & = \lim_{n \to \infty} \frac {\color{#3D99F6}{(n+1)^{m+1}}-n^{m+1}}{\sum_{k=1}^{n+1} k^m - \sum_{k=1}^{n} k^m} & \small \color{#3D99F6}{\text{By binomial expansion}} \\ & = \lim_{n \to \infty} \frac {\color{#3D99F6}{\left(n^{m+1}+(m+1)n^m + \frac {(m+1)m}2n^{m-1}+...+1\right)}-n^{m+1}}{(n+1)^m} \\ & = \lim_{n \to \infty} \frac {(m+1)n^m + \frac {(m+1)m}2n^{m-1}+ \frac {(m+1)m(m-1)}2n^{m-2}+...+1}{n^{m}+mn^{m-1} + \frac {m(m-1)}2 n^{m-2}+...+1} & \small \color{#3D99F6}{\text{Divide up and down by }n^m} \\ & = \lim_{n \to \infty} \frac {(m+1) + \frac {(m+1)m}2n^{-1}+ \frac {(m+1)m(m-1)}2n^{-2}+...+n^{-m}}{1+mn^{-1} + \frac {m(m-1)}2 n^{-2}+...+n^{-m}} \\ & = m+1 \end{aligned}$

For $m=4$ , the answer is $m+1 = \boxed{5}$ .