Sir care nu este monoton, dar are limita
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Cezar Lupu
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Sir care nu este monoton, dar are limita
Fie sirul \( (x_{n})_{n\geq 1} \) definit prin \( x_{n}=\prod_{k=1}^{n}\sin k \). Sa se arate ca sirul \( (x_{n}) \) nu este monoton, dar exista \( \lim_{n\to\infty}x_{n} \).
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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Marius Mainea
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\( \frac{x_{n+1}}{x_n}=\sin ({n+1}) \) care poate lua atat valori pozitive (in intervale de tipul \( (2k\pi,(2k+1)\pi) \)) si negative in rest.
Asadar sirul are un subsir cu termeni pozitivi si un subsir cu termeni negativi, deci nu poate fi monoton.
Deoarece in intervalele de tipul \( [-\frac{\pi}{6}+2p\pi,\frac{\pi}{6}+2p\pi] \) se afla un singur numar natural \( k_p \), atunci
\( |x_n|\leq \prod_{p=1}^{\alpha(n)}|\sin k_p|\leq\left(\frac{1}{2}\right)^{\alpha(n)}\to0 \) cand n tinde la infinit.
Asadar sirul are un subsir cu termeni pozitivi si un subsir cu termeni negativi, deci nu poate fi monoton.
Deoarece in intervalele de tipul \( [-\frac{\pi}{6}+2p\pi,\frac{\pi}{6}+2p\pi] \) se afla un singur numar natural \( k_p \), atunci
\( |x_n|\leq \prod_{p=1}^{\alpha(n)}|\sin k_p|\leq\left(\frac{1}{2}\right)^{\alpha(n)}\to0 \) cand n tinde la infinit.