Sirul a_n=n|sin n| nu are limita
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Sirul a_n=n|sin n| nu are limita
Sa se demonstreze ca sirul \( a_{n}=n | \sin n|,\ n\geq 1 \) nu are limita.
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.
a) Multimea \( \{|sin(n)|:\, n \in\mathbb{N}\} \) este densa in [0,1]. Exista deci un sir \( n_k\in\mathbb{N} \) cu \( |sin(n_k)| \to 1 \), deci \( a_{n_k}\to\infty \).
b) \( \pi \) fiind irational, exista o infinitate de perechi \( (p,q) \in\mathbb{N}^2, p,q > 2 \) astfel incat \( |\pi - p/q| < 1/q^2 \). Rezulta \( |sin(p)| = |sin(p - q\pi)|<1/q \), deci \( a_p < p/q < 4 \). Se obtine astfel un subsir marginit al lui \( (a_n) \).
Prin urmare sirul \( (a_n) \) nu are limita.
P.S. Ce se intampla cu sirul \( b_{n}=n^{2008}| \sin n| \)?
b) \( \pi \) fiind irational, exista o infinitate de perechi \( (p,q) \in\mathbb{N}^2, p,q > 2 \) astfel incat \( |\pi - p/q| < 1/q^2 \). Rezulta \( |sin(p)| = |sin(p - q\pi)|<1/q \), deci \( a_p < p/q < 4 \). Se obtine astfel un subsir marginit al lui \( (a_n) \).
Prin urmare sirul \( (a_n) \) nu are limita.
P.S. Ce se intampla cu sirul \( b_{n}=n^{2008}| \sin n| \)?