Question

Salut, am nevoie de ajutor la ex. 81

Answer 1

Explicație pas cu pas:

Fie termenul general al sirului :

$1-\frac{k}{n+1}$

$1-\frac{k}{n+1} = \frac{n+1}{n+1} -\frac{k}{n+1} = \frac{n-k+1}{n+1}$

sirul $a_{n}$ devine :

$a_{n} = \frac{n+1}{n+1}* \frac{n}{n+1} * \frac{n-1}{n+1} * ... * \frac{1}{n+1}$

$a_{n} = \frac{(n+1)!}{(n+1)^{n} }$

$\lim_{n \to \infty} \frac{a_{n+1} }{a_{n} } = \lim_{n \to \infty} \frac{(n+2)!}{(n+2)^{n+1} } * \frac{(n+1)^{n} }{(n+1)!} = \lim_{n \to \infty} \frac{(n+2)!}{(n+1)! } * \frac{(n+1)^{n} }{(n+2)^{n+1}} = \lim_{n \to \infty} \frac{(n+1)! * (n+2)}{(n+1)! } * \frac{(n+1)^{n} }{(n+2)^{n} * (n+2)} = \lim_{n \to \infty} (n+2)}{ } * \frac{(n+1)^{n} }{(n+2)^{n} * (n+2)} = \lim_{n \to \infty} (\frac{n+1}{n+2} ) ^{n}$

$\lim_{n \to \infty} (1+\frac{n+1}{n+2}-1 ) ^{n} = \lim_{n \to \infty} (1+\frac{n+1}{n+2}-\frac{n+2}{n+2} ) ^{n} = \lim_{n \to \infty} (1+\frac{-1}{n+2} )^{n} = \lim_{n \to \infty} ((1+\frac{-1}{n+2} )^{\frac{n+2}{-1} }) ^{\frac{-n}{n+2} } = e^{ \lim_{n \to \infty} a_n \frac{-n}{n+2} } = e^{-2} = \frac{1}{e^{2} }$