Fie \( (a_n)_{n\ge 0} \) un sir de numere reale pozitive definit prin:
\( a_n^2=n(a_{n-1}-a_n) , \forall n\ge1 \).
Sa se arate ca sirul \( x_n=(1+\frac{1}{2}+....+\frac{1}{n})a_n \) este convergent.
Gazeta Matematica
Sir convergent
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Marius Mainea
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andy crisan
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Daca \( \exists k\in\mathbb{N} \) a.i. \( a_{k}=0 \) \( \Rightarrow a_{n}=0\ (\forall)n\in\mathbb{N},\ n\geq k \). De unde concluzia.
Daca \( a_{n}\neq 0\ (\forall) n\in\mathbb{N} \Rightarrow a_{n-1}-a_{n}>0\ (\forall) n \in\mathbb{N}^*\Rightarrow \)\( (a_{n})_{n\geq0} \) este convergent fiind marginit superior de \( a_{0} \) si marginit inferior de \( 0 \).
Din relatia din enunt rezulta \( \frac{a_{n}^2}{n}=a_{n-1}-a_{n}\forall n\in\mathbb{N} \). Insumand toate "randurile" de la \( 1 \) la \( n \)
obtinem ca \( a_{1}-a_{n}=\sum_{k=1}^{n}{\frac{a_{k}^2}{k}}\geq a_{n}^2(1+\dots +\frac{1}{n})\Rightarrow \frac{a_{1}}{a_{n}^2}-\frac{1}{a_{n}}\to \infty \Rightarrow a_{n}\to 0 \).
Din teorema Cesaro-Stolz avem \( \lim_{n\to\infty}b_{n}=\lim_{n\to\infty}\frac{\frac{1}{n+1}}{\frac{a_{n-1}-a_{n}}{a_{n-1}a_{n}}}=\lim_{n\to\infty}{\frac{a_{n-1}}{a_{n}}}=1 \).
Daca \( a_{n}\neq 0\ (\forall) n\in\mathbb{N} \Rightarrow a_{n-1}-a_{n}>0\ (\forall) n \in\mathbb{N}^*\Rightarrow \)\( (a_{n})_{n\geq0} \) este convergent fiind marginit superior de \( a_{0} \) si marginit inferior de \( 0 \).
Din relatia din enunt rezulta \( \frac{a_{n}^2}{n}=a_{n-1}-a_{n}\forall n\in\mathbb{N} \). Insumand toate "randurile" de la \( 1 \) la \( n \)
obtinem ca \( a_{1}-a_{n}=\sum_{k=1}^{n}{\frac{a_{k}^2}{k}}\geq a_{n}^2(1+\dots +\frac{1}{n})\Rightarrow \frac{a_{1}}{a_{n}^2}-\frac{1}{a_{n}}\to \infty \Rightarrow a_{n}\to 0 \).
Din teorema Cesaro-Stolz avem \( \lim_{n\to\infty}b_{n}=\lim_{n\to\infty}\frac{\frac{1}{n+1}}{\frac{a_{n-1}-a_{n}}{a_{n-1}a_{n}}}=\lim_{n\to\infty}{\frac{a_{n-1}}{a_{n}}}=1 \).