Evaluate $$\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^n \sqrt{\frac{k^3}{n}}$$
At first, I think this is Riemann sum. But that was not.
If there is $\frac{1}{n^2}$ ( not $\frac{1}{n}$), that's correct but this case is $\frac{1}{n}$.
So I think I need lower bound and upper bound to squeeze this, but I couldn't find.
Thanks for help.
Your remark is correct, now split the limit as a product of two terms: as $n\to+\infty$, $$\frac{1}{n} \sum_{k=1}^n \sqrt{\frac{k^3}{n}}=\underbrace{n}_{\to+\infty}\cdot \underbrace{\frac{1}{n} \sum_{k=1}^n \left(\frac{k}{n}\right)^{3/2}}_{\to \int_0^1 x^{3/2}\,dx=2/5}\to +\infty.$$
$$\frac1n\sum_{k=1}^{n}\sqrt{\frac{k^3}{n}}=n\cdot\frac1n\sum_{k=1}^{n}\left(\frac kn\right)^{3/2}\to \left[\infty\times\int_0^1 x^{3/2}\,dx\right]=\infty$$
Note tha $$\sum_{k=1}^{n} k^p \sim \frac{n^{p+1}}{p+1},~ p>0.$$
Then $$f(n)= \sum_{k=1}^{n}\frac{k^{3/2}}{n^{3/2}} \sim \frac{2}{5} \frac{n^{5/2}}{n^{3/2}} \sim \frac{2}{5} n$$ so the given limit $$L= \lim_{n\rightarrow \infty} f(n)= \infty.$$
