Question

Folosind criteriul lui Cesaro Stolz, sa se calculeze limita sirului $C_{n}= \frac{n}{ 2^{n} }$

Answer 1

Daca sirul, bn, crescator nemarginit, atunci $\lim_{n \to \infty} \frac{ a_{n} }{ b_{n} }= \lim_{n \to \infty} \frac{ a_{n+1}- a_{n} }{ b_{n+1}-b _{n} }$
 Deci $\lim_{n \to \infty}c_n= \lim_{n \to \infty} \frac{n}{2^n}= \lim_{n \to \infty} \frac{(n+1)-n}{ 2^{n+1}-2^n }= \lim_{n \to \infty} \frac{1}{2^n(2-1)}=0$