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Question-73251




Question Number 73251 by aliesam last updated on 09/Nov/19
Answered by mind is power last updated on 09/Nov/19
E(u_n )≤u_n <E(u_n )+1=U_(n+1)   ⇒U_n is a creasing squances  U_(n+2) =E(E(u_n )+1)+1=E(u_n )+2=1+U_(n+1)   ⇒U_(n+2) −U_(n+1) =1  ⇒U_(n+1) −U_n =1 ∀n∈N^∗   ⇒Σ_(k=1) ^(n−1) (U_(k+1) −U_k )=Σ_(k=1) ^(n−1) 1  ⇒U_n −U_1 =n−1⇒U_n =(n−1)+U_1   (U_n /n)=(((n−1))/n)+(U_1 /n)→1
$$\mathrm{E}\left(\mathrm{u}_{\mathrm{n}} \right)\leqslant\mathrm{u}_{\mathrm{n}} <\mathrm{E}\left(\mathrm{u}_{\mathrm{n}} \right)+\mathrm{1}=\mathrm{U}_{\mathrm{n}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{U}_{\mathrm{n}} \mathrm{is}\:\mathrm{a}\:\mathrm{creasing}\:\mathrm{squances} \\ $$$$\mathrm{U}_{\mathrm{n}+\mathrm{2}} =\mathrm{E}\left(\mathrm{E}\left(\mathrm{u}_{\mathrm{n}} \right)+\mathrm{1}\right)+\mathrm{1}=\mathrm{E}\left(\mathrm{u}_{\mathrm{n}} \right)+\mathrm{2}=\mathrm{1}+\mathrm{U}_{\mathrm{n}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{U}_{\mathrm{n}+\mathrm{2}} −\mathrm{U}_{\mathrm{n}+\mathrm{1}} =\mathrm{1} \\ $$$$\Rightarrow\mathrm{U}_{\mathrm{n}+\mathrm{1}} −\mathrm{U}_{\mathrm{n}} =\mathrm{1}\:\forall\mathrm{n}\in\mathbb{N}^{\ast} \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\left(\mathrm{U}_{\mathrm{k}+\mathrm{1}} −\mathrm{U}_{\mathrm{k}} \right)=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\mathrm{1} \\ $$$$\Rightarrow\mathrm{U}_{\mathrm{n}} −\mathrm{U}_{\mathrm{1}} =\mathrm{n}−\mathrm{1}\Rightarrow\mathrm{U}_{\mathrm{n}} =\left(\mathrm{n}−\mathrm{1}\right)+\mathrm{U}_{\mathrm{1}} \\ $$$$\frac{\mathrm{U}_{\mathrm{n}} }{\mathrm{n}}=\frac{\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{n}}+\frac{\mathrm{U}_{\mathrm{1}} }{\mathrm{n}}\rightarrow\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

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