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Se considera sirul \( (a_{n})_{n\geq 1} \) definit prin \( a_{1}=1 \) si
\( a_{n+1}=\sqrt{a_{1}+a_{2}+\ldots +a_{n}}, \forall n\geq 1 \). Sa se calculeze \( \lim_{n\to\infty}\frac{a_{n}}{n} \).

\( (a_n) \) sir de numere pozitive.
Relatia din enunt e echivalenta cu \( a_{n+1}^2=a_n^2+a_n \), deci sirul e strict crescator si presupunand ca are limita \( l \) si trecand in relatie la limita, obtinem \( l=0 \), contradictie.
Deci \( \lim_{n\to\infty} a_n=\infty \)

\( \frac{a_{n+1}}{a_n} \rightarrow 1 \Rightarrow \frac{a_n}{a_{n+1}+a_n} \rightarrow \frac{1}{2} \)

Cu Cesaro Stolz, \( \lim_{n\to\infty} \frac{a_n}{n}=\lim_{n\to\infty}(a_{n+1}-a_n)= \lim_{n\to\infty} \frac{a_n}{a_n+a_{n+1}} =\frac{1}{2}. \)