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

(U_n )_(n∈N^∗ ) : U_n =Σ_(k=1) ^n (1/k) Show that ∀ n∈N^(∗ ) , U_(2n) −U_n ≥(1/2) . Deduct that lim_(n→+∞) U_n =+∞

$$\left({U}_{{n}} \right)_{{n}\in\mathbb{N}^{\ast} } \::\:{U}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\: \\ $$$${Show}\:{that}\:\forall\:{n}\in\mathbb{N}^{\ast\:} ,\:{U}_{\mathrm{2}{n}} −{U}_{{n}} \geqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$.\:{Deduct}\:{that}\:\underset{{n}\rightarrow+\infty} {{lim}}\:{U}_{{n}} =+\infty \\ $$

Answered by alephzero last updated on 07/Feb/22

2) lim_(n→+∞) U_n = Σ_(k=1) ^∞ (1/k) Test for converge/diverge ∫_1 ^∞ (dx/x) = +∞ ⇒ Σ_(k=1) ^∞ (1/k) diverges ⇒ lim_(n→+∞) U_n = +∞

$$\left.\mathrm{2}\right) \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}{U}_{{n}} \:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}} \\ $$$$\mathrm{Test}\:\mathrm{for}\:\mathrm{converge}/\mathrm{diverge} \\ $$$$\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{{dx}}{{x}}\:=\:+\infty\:\Rightarrow\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}}\:\mathrm{diverges} \\ $$$$\Rightarrow\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}{U}_{{n}} \:=\:+\infty \\ $$

Answered by Mathspace last updated on 08/Feb/22

we have u_(2n) −u_n =Σ_(k=1) ^(2n) (1/k) −Σ_(k=1) ^n (1/k)=Σ_(k=n+1) ^(2n) (1/k) or n+1≤k≤2n ⇒(1/(2n))≤(1/k)≤(1/(n+1)) ⇒Σ_(k=n+1) ^(2n) (1/(2n))≤Σ_(k=n+1) ^(2n) (1/k) ⇒ n×(1/(2n))≤u_(n+1) −u_n ⇒u_(n+1) −u_n ≥(1/2) if lim u_n =l ∈R ⇒lim u_(2n) =l (suite extrete de u_n ) ⇒lim (u_(2n) −u_n )=0 we get o≥(1/2) impossible so limu_n is infinite another way u_n =H_n and H_n ∼ln(n)+γ +o((1/n)) ⇒ lim H_n =limln(n)=+∞

$${we}\:{have}\:{u}_{\mathrm{2}{n}} −{u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}} \\ $$$$−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}=\sum_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}} \\ $$$${or}\:\:\:{n}+\mathrm{1}\leqslant{k}\leqslant\mathrm{2}{n}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}{n}}\leqslant\frac{\mathrm{1}}{{k}}\leqslant\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$\Rightarrow\sum_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \frac{\mathrm{1}}{\mathrm{2}{n}}\leqslant\sum_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}}\:\Rightarrow \\ $$$${n}×\frac{\mathrm{1}}{\mathrm{2}{n}}\leqslant{u}_{{n}+\mathrm{1}} −{u}_{{n}} \:\Rightarrow{u}_{{n}+\mathrm{1}} −{u}_{{n}} \geqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${if}\:{lim}\:{u}_{{n}} ={l}\:\:\in{R}\:\Rightarrow{lim}\:{u}_{\mathrm{2}{n}} ={l}\:\left({suite}\:\right. \\ $$$$\left.{extrete}\:{de}\:{u}_{{n}} \right)\:\Rightarrow{lim}\:\left({u}_{\mathrm{2}{n}} −{u}_{{n}} \right)=\mathrm{0} \\ $$$${we}\:{get}\:\:{o}\geqslant\frac{\mathrm{1}}{\mathrm{2}}\:{impossible}\:{so}\:{limu}_{{n}} \\ $$$${is}\:{infinite} \\ $$$${another}\:{way} \\ $$$${u}_{{n}} ={H}_{{n}} \:\:{and}\:{H}_{{n}} \sim{ln}\left({n}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow \\ $$$${lim}\:{H}_{{n}} ={limln}\left({n}\right)=+\infty \\ $$

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