A Fancy Limit

Limit can be written as: $\left(\displaystyle\lim_{n\to \infty}\dfrac{1}{n^2}\right)\times \left(\displaystyle\lim_{n\to \infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{k^5}{n^5}\right)$

Using Reimann Sums the second limit can be calculated (as $\frac 16$ ) while first limit is obviously $0$ . Hence overall limit is $0$ .

$\begin{aligned} \mathfrak {L} & = \lim_{n \to \infty} \frac{\small {\displaystyle \sum_{k=1}^n k^5}}{n^8} \\ & = \lim_{n \to \infty} \frac{\color{#3D99F6}{\frac{1}{12}(2n^6+6n^5+5n^4-n^2)}}{n^8} \quad \quad \quad \quad \small \color{#3D99F6}{\text{See Note}} \\ & = \lim_{n \to \infty} \left( \frac{1}{6n^2} + \frac{1}{2n^3} + \frac{5}{15n^4} - \frac{1}{12n^6} \right) \\ & = \boxed{0} \end{aligned}$

$\color{#3D99F6}{\text{Note:}}$ See equation (25) of Power Sum .