Computing limit based on upper and lower Riemann sums I have a question where it is asking me to compute the $$\lim_{n \to \infty} \frac{(\sqrt{1}+\sqrt{2}+...+\sqrt{n})}{(n\sqrt{n})}$$ based on the integral of sqrt(n) from the interval of $[0,n]$. Based on my previous answer, I was able to use the upper Riemann sum since it overlaps with the question nicely such that the numerator is equaled with the upper Riemann sum, but for the lower riemann sum, it’s kinda awkward since it goes from $\sqrt{1} + \sqrt{2} + ... + \sqrt{n-1}$ so any ideas on how I should proceed? Thanks :)

$\displaystyle\mathbf{Method}\mathbf{1}:$
$\displaystyle\lim_{n\rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\cdots +\sqrt{n}}{n\sqrt{n}}=\lim_{n\rightarrow \infty} \frac{1}{n}\left( \sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\cdots +\sqrt{\frac{n}{n}} \right) =\int_0^1{\sqrt{x}\text{d}x}=\frac{2}{3}x^{\frac{3}{2}}\mid_{0}^{1}=\frac{2}{3}$
$\displaystyle\mathbf{Method}\mathbf{2}:$
$\displaystyle\lim_{n\rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\cdots +\sqrt{n}}{n\sqrt{n}}=\lim_{n\rightarrow \infty} \frac{\sqrt{n}}{\sqrt{n^3}-\sqrt{\left( n-1 \right) ^3}}=\lim_{n\rightarrow \infty} \frac{\sqrt{n}\left( \sqrt{n^3}+\sqrt{\left( n-1 \right) ^3} \right)}{3n^2-3n+1}=\frac{2}{3}$
