Question

calculati limita din 2^n/n!

Answer 1

$\dfrac{2^n}{n!} = \dfrac{2\cdot 2}{1\cdot 2}\cdot \dfrac{2^{n-2}}{3\cdot 4\cdot \ldots \cdot n} < 2\cdot \left (\dfrac{2}{3}\right) ^{n-2}\\\text{Am obtinut astfel :}\\0<\dfrac{2^n}{n!}< 2\cdot \left (\dfrac{2}{3}\right) ^{n-2}\\\text{Din criteriul clestelui rezulta ca }\displaystyle\limit\lim _{n\to\infty} \dfrac{2^n}{n!} =0$