Calculate the limit using stolz theorem

Question

Can someone help me calculate $\lim_{x\to \infty} \frac{1+2\sqrt{2}+...+n\sqrt{n}}{n^2\sqrt{n}}$ using Stolz theorem? So far I've only managed to write it as $\lim_{x\to \infty} \frac{(n+1)\sqrt{n+1}}{(n+1)^2\sqrt{n+1}-n^2\sqrt{n}}$, I was wondering if I did it correctly and what would the next steps be. Thanks in advance!

Answer 1

Is Stolz compulsory? Otherwise you can rewrite the sum as a Riemann sum and calculate the integral:

$$\sum_{i=1}^n\left(\frac in \right)^{\frac 32}\cdot \frac 1n \stackrel{n\to \infty}{\longrightarrow} \int_0^1x^{\frac 32}dx = \frac 25$$

Answer 2

I think you want to use Stolz–Cesàro theorem, so I try to avoid a clear easy solution that consciously using integrals, as @trancelocation says.

It is trivial that denominator of $\frac{1\sqrt1+\cdots n\sqrt n}{n^2\sqrt n}$, $n^2\sqrt n$ is monotone increasing and diverges, so we can use $\cdot/\infty$ case of Stolz-Cesàro theorem.

Following is whole solution.

$$\lim_{n\to\infty}\frac{1\sqrt1+\cdots+n\sqrt n}{n^2\sqrt n}=\lim_{n\to\infty}\frac{n\sqrt n}{n^2\sqrt n-(n-1)^2\sqrt {n-1}}\\=\lim_{n\to\infty}\frac{n^{\frac32}(n^{\frac52}+(n-1)^\frac52)}{n^5-(n-1)^5}\\=\lim_{n\to\infty}\frac{1+(1-\frac1n)^\frac52}{5-\frac{10}n+\frac{10}{n^2}-\frac5{n^3}+\frac1{n^4}}\\=\frac25$$