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Question Number 165724 by mathocean1 last updated on 06/Feb/22

$${\left(U_n\right)}_{n\in {\mathbb{N}}^{\ast }}:U_n=\underset{k=1}{\sum ^n}\frac{1}{k}$$$$Showthat\mathrm{\forall }n\in {\mathbb{N}}^{\ast },U_{2n}-U_n⩾\frac{1}{2}$$$$.Deductthat\underset{n\to +\mathrm{\infty }}{lim}{U}_n=+\mathrm{\infty }$$

Answered by alephzero last updated on 07/Feb/22

$$2)$$$$\underset{n\to +\mathrm{\infty }}{\lim }{U}_n=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}$$$$\mathrm{Test}\mathrm{for}\mathrm{converge}/\mathrm{diverge}$$$$\underset{1}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{x}=+\mathrm{\infty }\Rightarrow \underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}\mathrm{diverges}$$$$\Rightarrow \underset{n\to +\mathrm{\infty }}{\lim }{U}_n=+\mathrm{\infty }$$

Answered by Mathspace last updated on 08/Feb/22

$$wehave{u}_{2n}-u_n=\sum _{k=1}^{2n}\frac{1}{k}$$$$-\sum _{k=1}^n\frac{1}{k}=\sum _{k=n+1}^{2n}\frac{1}{k}$$$$orn+1⩽k⩽2n\Rightarrow \frac{1}{2n}⩽\frac{1}{k}⩽\frac{1}{n+1}$$$$\Rightarrow \sum _{k=n+1}^{2n}\frac{1}{2n}⩽\sum _{k=n+1}^{2n}\frac{1}{k}\Rightarrow$$$$n\times \frac{1}{2n}⩽u_{n+1}-u_n\Rightarrow u_{n+1}-u_n⩾\frac{1}{2}$$$$iflim{u}_n=l\in R\Rightarrow lim{u}_{2n}=l(suite$$$$extretede{u}_n)\Rightarrow lim(u_{2n}-u_n)=0$$$$wegeto⩾\frac{1}{2}impossibleso{limu}_n$$$$isinfinite$$$$anotherway$$$$u_n=H_{n}and{H}_n\sim ln\left(n\right)+\gamma +o\left(\frac{1}{n}\right)\Rightarrow$$$$lim{H}_n=limln\left(n\right)=+\mathrm{\infty }$$

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