Xét hàm: $f(x)=\frac{2}{3}\sqrt{x^3},x\ge1$.
Theo định lý Lagrange trên đoạn $[n,n+1]$ tồn tại số $c\in(n,n+1)$ sao cho:
$f(n+1)-f(n)=f'(c)=\sqrt c$
Suy ra: $\sqrt n<f(n+1)-f(n)<\sqrt{n+1},\forall n\ge1$ .
Từ đó suy ra:
$\sum_{k=1}^n\sqrt k<\sum_{k=1}^n(f(k+1)-f(k))=f(n+1)-f(1)=\frac{2}{3}\sqrt{(n+1)^3}-\frac{2}{3}$
$\Rightarrow \frac{1}{\sqrt{n^3} }\sum\limits_{k = 1}^n {\sqrt k } < \frac{2}{3}\sqrt{\left ( \frac{n+1}{n} \right )^3 }- \frac{2}{3\sqrt{n^3} } $
Lại có:
$\sum_{k=1}^n\sqrt k=1+ \sum_{k=2}^n\sqrt k>1+\sum_{k=1}^{n-1}(f(k+1)-f(k))=1+f(n)-f(1)>\frac{2}{3}\sqrt{n^3}$
$\Rightarrow \frac{1}{\sqrt{n^3} }\sum\limits_{k = 1}^n {\sqrt k }>\frac{2}{3}$