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Question Number 133997 by rs4089 last updated on 26/Feb/21

$${lim}_{n\to \mathrm{\infty }}{\left(\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}}{n^2}\right)}^n$$

Answered by mathmax by abdo last updated on 27/Feb/21

$${\mathrm{U}}_{\mathrm{n}}={\left(\frac{1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{\mathrm{n}}}{{\mathrm{n}}^2}\right)}^{\mathrm{n}}\Rightarrow {\mathrm{U}}_{\mathrm{n}}={\left(\frac{{\mathrm{H}}_{\mathrm{n}}}{{\mathrm{n}}^2}\right)}^{\mathrm{n}}\mathrm{but}{\mathrm{H}}_{\mathrm{n}}=\mathrm{logn}+\gamma +\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)\Rightarrow$$$${\mathrm{U}}_{\mathrm{n}}\sim {\left(\frac{\mathrm{logn}+\gamma +\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)}{{\mathrm{n}}^2}\right)}^{\mathrm{n}}\Rightarrow {\mathrm{u}}_{\mathrm{n}}\sim {\mathrm{e}}^{\mathrm{nlog}\left(\frac{\mathrm{logn}+\gamma +\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)}{{\mathrm{n}}^2}\right)}$$$$\mathrm{we}\mathrm{have}\frac{\mathrm{logn}+\gamma +\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)}{{\mathrm{n}}^2}\to 0^{+}\Rightarrow \mathrm{nlog}(.....)\to -\mathrm{\infty }\Rightarrow$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}={\mathrm{e}}^{-\mathrm{\infty }}=0$$

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