What is the $\lim _{n \to \infty} \sum_{k=0}^{n} \frac{\sqrt k }{n^{\frac32}}$? What is $$\lim _{n\to \infty }\left(\frac{\sum _{k=1}^n\sqrt{k}\:}{n^{\frac32}}\right)$$
Wolframalpha tells me its 2/3 but I don't understand why.
What I do know is that:
$$\lim _{n\to \infty }\left(\frac{\sum _{k=1}^n\sqrt{k}\:}{n^{\frac32}}\right) < \lim _{n\to \infty }\left(\frac{n\sqrt{n}\:}{n^{\frac32}}\right) =1$$
 A: The key-word is "Riemann sum" :$$\frac{\sum _{k=1}^n\sqrt{k}}{n^{1.5}}=\frac{1}{n}\sum _{k=1}^n\sqrt{\frac{k}{n}} \rightarrow_{n \rightarrow +\infty} \int_0^1 \sqrt{x} \ dx = \frac{2}{3}$$
A: By Stolz–Cesàro theorem we obtain
$$\frac{\sum _{k=1}^{n+1}\sqrt{k}-\sum _{k=1}^n\sqrt{k}}{\sqrt{(n+1)^3}-\sqrt{n^3}}=\frac{\sqrt{n+1}}{\sqrt{(n+1)^3}-\sqrt{n^3}}=$$
$$=\frac{\sqrt{n+1}}{\sqrt{(n+1)^3}-\sqrt{n^3}}\cdot\frac{\sqrt{(n+1)^3}+\sqrt{n^3}}{\sqrt{(n+1)^3}+\sqrt{n^3}}=\frac{\sqrt{n+1}\left(\sqrt{(n+1)^3}+\sqrt{n^3}\right)}{3n^2+3n+1} \to \frac23$$
A: If you know about generalized harmonic numbers, we can do amazing things.
Let
$$S_n=\frac 1 {n^\frac 32} \sum_{k=0}^n \sqrt k=\frac {H_n^{\left(-\frac{1}{2}\right)}} {n^\frac 32}$$ The asymptotics being
$$H_n^{\left(-\frac{1}{2}\right)}=\frac{2 n^{3/2}}{3}+\frac{\sqrt{n}}{2}+\zeta
   \left(-\frac{1}{2}\right)+O\left(\frac{1}{n^{3/2}}\right)$$ then
$$S_n=\frac{2}{3}+\frac{1}{2 n}+\cdots$$
A: The Mean Value Theorem says
$$
\tfrac32k^{1/2}\le\overbrace{(k+1)^{3/2}-k^{3/2}}^{\frac32\kappa^{1/2}\text{ for }k\lt\kappa\lt k+1}\le\tfrac32(k+1)^{1/2}
$$
Summing, we get
$$
\frac32\sum_{k=0}^{n-1}k^{1/2}\le n^{3/2}\le\frac32\sum_{k=1}^nk^{1/2}
$$
or equivalently
$$
\frac23n^{3/2}\le\sum_{k=1}^nk^{1/2}\le\frac23n^{3/2}+n^{1/2}
$$
Divide by $n^{3/2}$ and take the limit.
