Đáp án:
Giải thích các bước giải:
ta có
$\frac{1}{\sqrt[]{n}(n+1)}=\frac{\sqrt[]{n}}{n(n+1)}$
$=\sqrt[]{n}(\frac{1}{n}-\frac{1}{n+1})$
$=\sqrt[]{n}(\frac{1}{\sqrt[]{n}}-\frac{1}{\sqrt[]{n+1}})(\frac{1}{\sqrt[]{n}}+\frac{1}{\sqrt[]{n+1}})$
$ ≤\sqrt[]{n}(\frac{1}{\sqrt[]{n}}-\frac{1}{\sqrt[]{n+1}})(\frac{1}{\sqrt[]{n}}+\frac{1}{\sqrt[]{n}})$
$=2(\frac{1}{\sqrt[]{n}}-\frac{1}{\sqrt[]{n+1}})$
suy ra
biểu thức đã cho
$≤2(\frac{1}{\sqrt[]{1}}-\frac{1}{\sqrt[]{2}}+\frac{1}{\sqrt[]{2}}-\frac{1}{\sqrt[]{3}}...+\frac{1}{\sqrt[]{2019}}-\frac{1}{\sqrt[]{2020}})$
$=2(1-\frac{1}{\sqrt[]{2020}})$
$<2(1-\frac{1}{44})=\frac{88}{45}$
