Integration Upper Sum

Question

I have been given a real sequence $\{a_n\}_{n=1}^\infty$ by $$ a_n=\frac{1}{n} \sum_{j=1}^n \sqrt{\frac{j}{n}}$$ and i have to argue that the sequence is an upper sum for the integral $$ \int_0^1 \sqrt{x} \ dx$$ I am not sure how to do this though. I have looked at the definition for upper sums given by: $$ U(D)=\sum_{i=1}^n G_i(x_i-x_{i-1})$$ where $$ G_i=\sup \{f(x) \ | \ x\in[x_{i-1},x_i]\}$$ However, i cannot see how this is supposed to help me here..

Answer 1

This is due to the fact that $f(x) = \sqrt{x}$ is monotone increasing, therefore the upper Riemann sum: $$ \frac{1-0}{n}\bigg(\sqrt{\frac{0+1}{n}} + \sqrt{\frac{0+2}{n}} + \ldots + \sqrt{\frac{n}{n}} \bigg)=\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{k}{n}} $$ is the upper bound on the integral $$ \int_{0}^{1}\sqrt{x}dx $$

Answer 2

You could work the summation using generalized harmonic numbers $$a_n=\frac{1}{n} \sum_{j=1}^n \sqrt{\frac{j}{n}}=\frac{1}{n^{3/2}}\sum_{j=1}^n \sqrt{j}=\frac{1}{n^{3/2}}H_n^{\left(-\frac{1}{2}\right)}$$ Using asymptotics $$H_n^{\left(-\frac{1}{2}\right)}=\frac{2 n^{3/2}}{3}+\frac{n^{1/2}}{2}+\zeta \left(-\frac{1}{2}\right)+O\left(\frac{1}{n^{3/2}}\right)$$ $$\frac{1}{n^{3/2}}H_n^{\left(-\frac{1}{2}\right)}=\frac{2}{3}+\frac{1}{2 n}+O\left(\frac{1}{n^{3/2}}\right)$$ which shows the limit and how it is approached.