28437: Difference between revisions

Revision as of 17:07, 8 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 $\lim _{n\to \infty }({\frac {a_{n+1}}{a_{n}}}-1)\cdot e^{a_{n}}.$

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

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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>
+</br></br>
+'''Soluție:'''
 </br>
-'' Fie șirul '' <math> ((a_n))_{n \geq 1} </math> '' cu termenii strict pozitivi, dat de relația <math> a_{n+1} </math>
+Pentru orice <math> {n \geq 2} </math> avem <math>a_n = ln(a_1 + a_2 + ... + a_{n-1})
+</math>, deci <math>a_n = a_1 + a_2 + ... + a_{n-1} = e^{a_n}</math>. Rezultă că pentru orice <math> {n \geq 2} </math> are loc
+</br>
+<math display = "block">a_{n+1}=ln(e^{a_n} + a_n).</math>
+</br>
+Deoarece <math> a_{n+1} - a_n = ln(e^{a_n} + a_n) - ln (e^{a_n} \ge 0) </math> pentru orice <math>{n \geq 2}</math> deducem că șirul <math> (a_n)_{n \geq 2} </math> este strict crescător.
+</br>
+Dacă șirul <math> (a_n)_{n \geq 2} </math> este mărginit superior, atunci <math> (a_n)_{n \geq 2} </math> este convergent cu <math>\lim_{{n \to \infty}} (a_n) = a \in (0, \infty). </math> Trecând la limită în relația (1), obținem