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Question Number 144929 by gsk2684 last updated on 30/Jun/21

$$\mathrm{find}\mathrm{the}\mathrm{value}$$$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{(1+\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{\mathrm{n}}}{{\mathrm{n}}^2})}^{\mathrm{n}}$$

Answered by mathmax by abdo last updated on 30/Jun/21

$${\mathrm{A}}_{\mathrm{n}}={(1+\frac{{\mathrm{H}}_{\mathrm{n}}}{{\mathrm{n}}^2})}^{\mathrm{n}}\Rightarrow {\mathrm{A}}_{\mathrm{n}}={\mathrm{e}}^{\mathrm{nlog}(1+\frac{{\mathrm{H}}_{\mathrm{n}}}{{\mathrm{n}}^2})}$$$${\mathrm{H}}_{\mathrm{n}}\sim \mathrm{logn}+\gamma +\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)\Rightarrow \log (1+\frac{{\mathrm{H}}_{\mathrm{n}}}{{\mathrm{n}}^2})\sim \log (1+\frac{\mathrm{logn}+\gamma +\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)}{{\mathrm{n}}^2})$$$$\sim \frac{\log \left(\mathrm{n}\right)+\gamma +\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)}{{\mathrm{n}}^2}\Rightarrow \mathrm{nlog}(1+\frac{{\mathrm{H}}_{\mathrm{n}}}{{\mathrm{n}}^2})\sim \frac{\mathrm{logn}+\gamma +\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)}{\mathrm{n}}\to 0(\mathrm{n}\to +\mathrm{\infty })$$$$\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}=1$$

Commented by gsk2684 last updated on 30/Jun/21

$$\mathrm{thank}\mathrm{you}$$

Commented by mathmax by abdo last updated on 30/Jun/21

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

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