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Question Number 71664 by mathmax by abdo last updated on 18/Oct/19

$$findnatureofthesequence{U}_n=\frac{1}{n}{\left(\sum _{k=1}^n\frac{1}{k}\right)}^2$$

Commented by mathmax by abdo last updated on 18/Oct/19

$$wehave\sum _{i=1}^n{x}_i^2⩾\frac{1}{n}{\left(\sum _{i=1}^n{x}_i\right)}^2\left(resultproved\right)$$$$let{x}_i=\frac{1}{i}\Rightarrow \sum _{i=1}^n\frac{1}{i^2}⩾\frac{1}{n}{\left(\sum _{i=1}^i\frac{1}{i}\right)}^2\Rightarrow 0<U_n⩽\sum _{i=1}^n\frac{1}{i^2}\Rightarrow$$$$0⩽{lim}_{n\to +\mathrm{\infty }}{U}_n⩽\frac{{\pi }^2}{6}thesequenc{U}_{n}converges.$$$$anotherwaywehave\sum _{k=1}^n\frac{1}{k}=H_n=ln\left(n\right)+\gamma +o\left(\frac{1}{n}\right)\Rightarrow$$$$\Rightarrow H_n\sim ln\left(n\right)+\gamma \Rightarrow H_n^2\sim {ln}^2\left(n\right)+2\gamma ln\left(n\right)+{\gamma }^2\sim {ln}^2\left(n\right)(n\to +\mathrm{\infty })\Rightarrow$$$$U_n\sim \frac{{ln}^2\left(n\right)}{n}wehave{lim}_{n\to +\mathrm{\infty }}\frac{{ln}^2\left(e^n\right)}{e^n}={lim}_{n\to +\mathrm{\infty }}{n}^2{e}^{-n}=0\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{U}_n=0$$

Answered by mind is power last updated on 18/Oct/19

$$\mathrm{\forall }\mathrm{n}\in {\mathbb{N}}^{\ast }\mathrm{we}\mathrm{have}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}⩾1$$$$\Rightarrow \mathrm{Un}⩾\frac{1}{\mathrm{n}}$$$$\Rightarrow \mathrm{\Sigma }{\mathrm{U}}_{\mathrm{n}}⩾\mathrm{\Sigma }\frac{1}{\mathrm{n}}\to +\mathrm{\infty }$$

Commented by mathmax by abdo last updated on 18/Oct/19

$$sirthequestionisfindthenatureofthesequencenottheserie...$$

Commented by mind is power last updated on 18/Oct/19

$$\mathrm{ok}\mathrm{sorry}$$$$\mathrm{for}\mathrm{the}\mathrm{sequence}$$$$\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{k}}\backsim \ln \left(\mathrm{n}\right)$$$$\Rightarrow \mathrm{Un}\backsim \frac{{\ln }^2\left(\mathrm{n}\right)}{\mathrm{n}}\to 0$$$$\mathrm{un}\to 0$$

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