Limit with Stolz-Cesàro theorem: $\lim\limits_{n\to \infty} \frac{1+2\sqrt2+3\sqrt3+\ldots+n\sqrt n}{n^2 \sqrt{n}}$ $$\lim_{n\to \infty} \frac{1+2\sqrt2+3\sqrt3+\ldots+n\sqrt n}{n^2 \sqrt{n}}= \text{?}$$
Book has no answers.
It's on Stolz-Cesàro theorem lesson, if that helps. Can't find a solution.

This is the direct application of the Stolz-Cesàro theorem
$$
\begin{align}
&\lim_{n\to \infty} \frac{1+2\sqrt{2}+...+n\sqrt{n}}{n^2 \sqrt{n}}\\
=&\lim_{n\to \infty} \frac{(1+2\sqrt{2}+...+(n+1)\sqrt{n+1})-(1+2\sqrt{2}+...+n\sqrt{n})}{(n+1)^2 \sqrt{n+1}-n^2\sqrt{n}}\\
=&\lim_{n\to \infty} \frac{(n+1)\sqrt{n+1}}{(n+1)^2 \sqrt{n+1}-n^2\sqrt{n}}\\
=&\lim_{n\to \infty} \frac{(n+1)\sqrt{n+1}((n+1)^2 \sqrt{n+1}+n^2\sqrt{n})}{((n+1)^2 \sqrt{n+1}-n^2\sqrt{n})((n+1)^2 \sqrt{n+1}+n^2\sqrt{n})}\\
=&\lim_{n\to \infty} \frac{(n+1)\sqrt{n+1}((n+1)^2 \sqrt{n+1}+n^2\sqrt{n})}{(n+1)^5-n^5}\\
=&\lim_{n\to \infty} \frac{(n+1)^4+(n+1)n^2\sqrt{n(n+1)}}{5n^4+10n^3+10n^2+5n+1}\\
=&\lim_{n\to \infty} \frac{(1+n^{-1})^4+(1+n^{-1})\sqrt{1+n^{-1}}}{5+10n^{-1}+10n^{-2}+5n^{-3}+n^{-4}}\\
=&\frac{2}{5}
\end{align}
$$
