Expressing $\lim\limits_{n\to\infty} \frac{1}{n^{3/2}}\sum\limits_{i=1}^n \sqrt i$ as an integral

Question

Problem

I have the following sum:

$$\lim\limits_{n\to\infty} \frac{1}{n^{3/2}}\sum\limits_{i=1}^n \sqrt i$$

My thoughts

This is in the chapter about integrals, and I recognize that this looks similar to the type of limit/sum that can be expressed as a definite integral. I've even ventured an "educated guess" that this would equal $$\int_0^1\sqrt x\mathrm dx$$ but the problem I have is that this is, as stated, just a guess. I don't know how to show that this is correct.

More specifically, my problem lies with the $\displaystyle\frac{1}{n^{3/2}}$ part. I can't see how I'm meant to work that in.

Any help appreciated!

Answer 1

Hint: \begin{align} \frac{1}{n^{3/2}}\sum\limits_{i=1}^n \sqrt i=\frac{1}{n}\sum\limits_{i=1}^n \sqrt {\frac i n}\\ \end{align}

Answer 2

The right Riemann sum (see here for more details) is defined by $$ \Delta x[f(a+\Delta x)+f(a+2\Delta x)+\ldots+f(b)]. $$ Set $\Delta x=n^{-1}$, $f(x)=\sqrt{x}$ for each $x\ge0$, $a=0$ and $b=1$. Then the right Riemann sum becomes $$ \frac1n\sum_{i=1}^n\sqrt{\frac i{n}}=\frac1{n^{3/2}}\sum_{i=1}^n\sqrt{i}. $$ Since the function $x\mapsto\sqrt{x}$ is Riemann integrable, the Riemann sum converves to the integral and we have that $$ \lim_{n\to\infty}\frac1{n^{3/2}}\sum_{i=1}^n\sqrt{i} =\lim_{n\to\infty}\frac1n\sum_{i=1}^n\sqrt{\frac{i}{n}} =\int_0^1\sqrt x\mathrm dx. $$

Answer 3

$\frac{1}{n^{3/2}}\sum\limits_{i=1}^n \sqrt i=\frac{1}{n}\sum\limits_{i=1}^n \sqrt{ i/n}=\frac{1}{n}\sum\limits_{i=1}^n f(i/n)$ with $f(x)=\sqrt{x}$

Hence $\frac{1}{n^{3/2}}\sum\limits_{i=1}^n \sqrt i \to \int_0^1\sqrt x\mathrm dx$