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Question Number 148151 by puissant last updated on 25/Jul/21

Answered by Olaf_Thorendsen last updated on 25/Jul/21

$$1.$$$${\left(u_n\right)}_{n\in {\mathbb{N}}^{\ast }}\mathrm{est}\mathrm{de}\mathrm{type}\mathcal{M}.$$$$\mathrm{Donc}\mathrm{\forall }n\in {\mathbb{N}}^{\ast },u_n=\frac{1}{n}\underset{k=1}{\sum ^n}{u}_{n+k}$$$$\mathrm{alors}{u}_n-\mathrm{C}=\left(\frac{1}{n}\underset{k=1}{\sum ^n}{u}_{n+k}\right)-\mathrm{C}$$$$u_n-\mathrm{C}=\frac{1}{n}\underset{k=1}{\sum ^n}{u}_{n+k}-\frac{1}{n}\underset{k=1}{\sum ^n}\mathrm{C}$$$$u_n-\mathrm{C}=\frac{1}{n}\underset{k=1}{\sum ^n}(u_{n+k}-\mathrm{C})$$$$\mathrm{et}\mathrm{donc}\mathrm{la}\mathrm{suite}{(u_n-\mathrm{C})}_{n\in {\mathbb{N}}^{\ast }}\mathrm{est}\mathrm{de}\mathrm{type}\mathcal{M}.$$$$$$$$2.$$$${\left(u_n\right)}_{n\in {\mathbb{N}}^{\ast }}\mathrm{est}\mathrm{de}\mathrm{type}\mathcal{M}.$$$$\mathrm{Donc}\mathrm{\forall }n\in {\mathbb{N}}^{\ast },u_n=\frac{1}{n}\underset{k=1}{\sum ^n}{u}_{n+k}$$$$\mathrm{Si}\mathrm{la}\mathrm{suite}\left(u_n\right)\mathrm{est}\mathrm{croissante}\mathrm{alors}\mathrm{les}$$$$\mathrm{termes}{u}_{n+k}\mathrm{sont}\mathrm{tous}\mathrm{plus}\mathrm{grands}\mathrm{que}$$$$u_n\mathrm{et}\mathrm{donc}{u}_n=\frac{1}{n}\underset{k=1}{\sum ^n}{u}_{n+k}⩽\frac{1}{n}\left({nu}_n\right)$$$$\mathrm{Soit}\mathrm{\forall }n\in {\mathbb{N}}^{\ast },u_n⩽u_n\mathrm{ce}\mathrm{qui}\mathrm{est}$$$$\mathrm{possible}\mathrm{uniquement}\mathrm{en}\mathrm{cas}{\mathrm{d}}^{\prime }\mathrm{egalite}$$$$\mathrm{stricte}.\mathrm{Si}\mathrm{un}\mathrm{seul}\mathrm{terme}\mathrm{de}\mathrm{la}\mathrm{somme}$$$$\mathrm{etait}\mathrm{strictement}\mathrm{superieur}\mathrm{a}{u}_n\mathrm{alors}$$$$\mathrm{on}\mathrm{aboutirait}\mathrm{a}\mathrm{la}\mathrm{contradiction}$$$$u_n<u_n\mathrm{ce}\mathrm{qui}\mathrm{est}\mathrm{absurde}.$$$$\mathrm{En}\mathrm{conclusion},\mathrm{toute}\mathrm{suite}\mathrm{croissante}$$$$\mathrm{au}\mathrm{sens}\mathrm{large}\mathrm{de}\mathcal{M}\mathrm{est}\mathrm{constante}.$$$$$$$$3.$$$${\left(u_n\right)}_{n\in {\mathbb{N}}^{\ast }}\mathrm{est}\mathrm{de}\mathrm{type}\mathcal{M}.$$$$\mathrm{Donc}\mathrm{\forall }n\in {\mathbb{N}}^{\ast },u_n=\frac{1}{n}\underset{k=1}{\sum ^n}{u}_{n+k}\left(1\right)$$$$\mathrm{Si}\mathrm{les}{u}_n\mathrm{sont}\mathrm{de}\mathrm{la}\mathrm{forme}{an}^2+bn+c$$$$\left(1\right):{an}^2+bn+c=\frac{1}{n}\underset{k=1}{\sum ^n}(a{(n+k)}^2+b(n+k)+c)$$$${an}^2+bn+c=\frac{1}{n}\underset{k=1}{\sum ^n}({ak}^2+k(2an+b)+{an}^2+bn+c)$$$${an}^2+bn+c=\frac{1}{n}\underset{k=1}{\sum ^n}({ak}^2+k(2an+b))+{an}^2+bn+c$$$$\Rightarrow \underset{k=1}{\sum ^n}({ak}^2+k(2an+b))=0\left(2\right)$$$$\left(2\right){\mathrm{n}}^{\prime }\mathrm{est}\mathrm{possible}\mathrm{que}\mathrm{quand}a=b=0.$$

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