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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 <math> a = ln(e^{a_n} + a)</math> de unde <math> a = 0 </math>, absurd! Prin urmare, șirul <math>((a_n)_{n \geq 1}</math> este crescător și nemărginit superior, deci <math>\lim_{{n \to \infty}} a_n =\infty</math>. | 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 <math> a = ln(e^{a_n} + a)</math> de unde <math> a = 0 </math>, absurd! Prin urmare, șirul <math>((a_n)_{n \geq 1}</math> este crescător și nemărginit superior, deci <math>\lim_{{n \to \infty}} a_n =\infty</math>. | ||
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Atunci <math>\lim_{{n \to \infty}}(\frac{a_{n+1}}{a_n}-1) \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}</math> | Atunci <math>\lim_{{n \to \infty}}(\frac{a_{n+1}}{a_n}-1) \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(1+\frac{a_n}{e^{a_n})}{\frac{a_n}{e^{a_n}}=1</math> deoarece din | ||
Revision as of 19:20, 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 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+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
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}}(\frac{a_{n+1}}{a_n}-1) \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(1+\frac{a_n}{e^{a_n})}{\frac{a_n}{e^{a_n}}=1} deoarece din