Proving $\sqrt 1+\sqrt 2+..+\sqrt n \approx\frac{2}{3} n^{\frac{3}{2}}$ asymptotically

Question

Let $n$ be a positive intiger, prove this asymptotic formula for large $n$

$$\sqrt1+\sqrt2+\cdots +\sqrt n=\frac{2}{3}n^{\frac{3}{2}}+\text{lower order}$$ using a Riemann sum.

Answer 1

We can obtain your approximation as follows: $$\sum\limits_{k=1}^nk^\frac12=n^\frac32\sum\limits_{k=0}^n\left(\frac kn\right)^\frac12\frac1n=n^{\frac32}g\left(n\right)$$ where $g\left(n\right)=\sum\limits_{k=0}^n\left(\frac kn\right)^\frac12\frac1n$. Now for large $n$, we have $$\lim\limits_{n\to\infty}g\left(n\right)=\lim\limits_{n\to\infty}\sum_{k=0}^n\left(\frac kn\right)^\frac12\frac1n=\int_0^1x^\frac12\mathrm dx=\frac23$$ So for large $n$, we have $g\left(n\right)=\frac23+\left(\text{things vanish at large }n\right)$. Thus for large $n$, $$\sum_{k=1}^nk^\frac12=\left(\frac23+\text{things that vanish at large }n\right)n^\frac32\\ =\frac23n^\frac32+\text{lower order}$$