Question Number 179950 by mathlove last updated on 04/Nov/22
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)+\left(\mathrm{1}+\frac{\mathrm{2}}{{x}}\right)^{\frac{\mathrm{2}}{\mathrm{2}}} +\left(\mathrm{1}+\frac{\mathrm{3}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} +\centerdot\centerdot\centerdot\centerdot+\left(\mathrm{1}+\frac{{x}}{{x}}\right)^{\frac{\mathrm{1}}{{x}}} \right]=? \\ $$
Commented by mahdipoor last updated on 04/Nov/22
$$ \\ $$$${x}>{n}>\mathrm{0}\Rightarrow\mathrm{1}+\frac{{n}}{{x}}>\mathrm{1}\Rightarrow\left(\mathrm{1}+\frac{{n}}{{x}}\right)^{\mathrm{1}/{n}} >\mathrm{1}^{\mathrm{1}/{n}} \\ $$$$\Rightarrow{A}\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{{x}} {\sum}}\left(\mathrm{1}+\frac{{n}}{{x}}\right)^{\mathrm{1}/{n}} >\left[\underset{{n}=\mathrm{1}} {\overset{{x}} {\sum}}\left(\mathrm{1}^{\mathrm{1}/{n}} =\mathrm{1}\right)\right]={x} \\ $$$$\Rightarrow\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({A}\left({x}\right)\right)=\infty \\ $$