28437: Difference between revisions

Revision as of 14:18, 11 November 2023

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 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 = \ln(a_1 + a_2 + ... + a_{n-1}) } , 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 a_n = a_1 + a_2 + ... + a_{n-1} = e^{a_n}} . Rezultă că 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} } are loc
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(e^{a_n} + a_n).}
Deoarece 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} - a_n = \ln(e^{a_n} + a_n) - \ln (e^{a_n} \ge 0) } 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}} deducem că 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 2} } este strict crescător.
Dacă 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 2} } este mărginit superior, 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 (a_n)_{n \geq 2} } este convergent cu 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) = a \in (0, \infty). } Trecând la limită în relația (1), obținem 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 = \ln(e^{a_n} + a)} de unde 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 = 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}

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 '''28437 (Nicolae Mușuroaia)'''
 </br></br>
-'' Fie șirul '' <math> (a_n)_{n \geq 1} </math> '' cu termenii strict pozitivi, dat de relația'' <math> a_{n+1}=\ln(a_1 + a_2 + ... + a_n), n \geq 1. </math>'' Determinați ''<math>\lim_{{n \to \infty}} (\frac{a_{n+1}}{a_n}-1) \cdot e^{a_n}. </math>
+'' Fie șirul '' <math> (a_n)_{n \geq 1} </math> '' cu termenii strict pozitivi, dat de relația'' <math> a_{n+1}=\ln(a_1 + a_2 + ... + a_n), n \geq 1. </math>'' Determinați ''<math>\lim_{{n \to \infty}} \left(\frac{a_{n+1}}{a_n}-1\right) \cdot e^{a_n}. </math>
 </br></br>
 '''Soluție:'''