20) [tex]\sqrt[n]{(x+1)(x+2)...(x+n)}-x\leqslant \frac{nx+1+...+n}{n}-x=\frac{1+...+n}{n}=\frac{\frac{n(n+1)}{2}}{n}=\frac{n+1}{2}[/tex]
[tex]\lim_{x\rightarrow +\infty }\left (\sqrt[n]{(x+1)(x+2)...(x+n)}-x \right )=\lim_{x\rightarrow +\infty }(\frac{n+1}{2})[/tex]
mà [tex]n>=1[/tex] => [tex]\lim_{x\rightarrow +\infty }(\frac{n+1}{2})=+\infty[/tex]
=> [tex]\lim_{x\rightarrow +\infty }\left (\sqrt[n]{(x+1)(x+2)...(x+n)}-x \right )=+\infty[/tex]