28437

28437 (Nicolae Mușuroaia)

Fie șirul Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_n)_{n \geq 1} } cu termenii strict pozitivi, dat de relația Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{n+1}=\ln(a_1 + a_2 + ... + a_n), n \geq 1. } Determinați Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{{n \to \infty}} \left(\frac{a_{n+1}}{a_n}-1\right) \cdot e^{a_n}. }

Soluție:
Pentru orice Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {n \geq 2} } avem $a_{n}=\ln(a_{1}+a_{2}+...+a_{n-1})$, deci $a_{n}=a_{1}+a_{2}+...+a_{n-1}=e^{a_{n}}$. Rezultă că pentru orice ${n\geq 2}$ are loc

$$a_{n+1}=\ln(e^{a_{n}}+a_{n}).$$

Deoarece $a_{n+1}-a_{n}=\ln(e^{a_{n}}+a_{n})-\ln(e^{a_{n}}\geq 0)$ pentru orice ${n\geq 2}$ deducem că șirul $(a_{n})_{n\geq 2}$ este strict crescător.
Dacă șirul $(a_{n})_{n\geq 2}$ este mărginit superior, atunci $(a_{n})_{n\geq 2}$ este convergent cu $\lim _{n\to \infty }(a_{n})=a\in (0,\infty ).$ Trecând la limită în relația (1), obținem $a=\ln(e^{a_{n}}+a)$ de unde $a=0$, absurd! Prin urmare, șirul Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ((a_n)_{n \geq 1}} este crescător și nemărginit superior, deci Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{{n \to \infty}} a_n =\infty} .
Atunci Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{{n \to \infty}} \left(\frac{a_{n+1}}{a_n}-1\right) \cdot e^{a_n} = \lim_{{n \to \infty}} \frac{\ln(e^{a_n} + a_n) - \ln(e^{a_n})}{a_n} \cdot e^{a_n} = \lim_{{n \to \infty}} \frac{\ln\left(1+\frac{a_n}{e^{a_n}}\right)}{\frac{a_n}{e^{a_n}}} = 1} deoarece din Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{{n \to \infty}} a_n =\infty } rezultă că Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{{n \to \infty}} \frac{a_n}{e^{a_n}}=0}