Question

studiati monotonia si marginea sirului

n! supra (1+1 la patrat)(1+2 la patrat)...(1+n la patrat)

Answer 1

$\displaystyle x_n= \frac{n!}{\prod\limits_{k=1}^n(1+k^2)}. \\ \\ Evident,~toti~termenii~sunt~pozitivi. \\ \\ \\ Avem~\frac{x_{n+1}}{x_n}= \frac{n+1}{1+(n+1)^2} \le \frac{1}{2}\ \textless \ 1.~Am~folosit~faptul~ca \\ \\ t^2+1 \ge 2t~(care~este~adevarat~din~inegalitate~mediilor,~sau~ \\ \\ observand~ca~t^2+1 \ge 2t \Leftrightarrow (t-1)^2 \ge 0.) \\ \\ Din~ \frac{x_{n+1}}{x_n}\ \textless \ 1,~rezulta~x_{n+1}\ \textless \ x_n.~Deci~sirul~este~strict \\ \\ descrescator.$

$\displaystyle Folosind~faptul~ca~ \frac{k}{k^2+1} \le \frac{1}{2}~(demonstrat~anterior),~obtinem: \\ \\ x_n= \frac{n!}{\prod\limits_{k=1}^n(1+k^2)}= \prod\limits_{k=1}^n \frac{k}{k^2+1} \le \left( \frac{1}{2} \right)^n. \\ \\ Din~0\ \textless \ x_n \le \left(\frac{1}{2} \right)^n,~obtinem~ \lim_{n \to \infty}x_n=0.$

Answer 2

Am atasat rezolvarea.

6174bb77864bb1d333e42219b848756b.pdf