Question

demonstrati ca sirul definit prin formula
$an = \frac{2n + 9}{n + 3}$
este descrescator

Answer 1

$\text{Cel mai simplu ar fi sa demosntram ca}\ \dfrac{a_{n+1}}{a_n}<1\\ \dfrac{\frac{2(n+1)+9}{n+1+3}}{\frac{2n+9}{n+3}}=\dfrac{2n+11}{n+4}\cdot \dfrac{n+3}{2n+9}= \dfrac{2n^2+6n+11n+33}{2n^2+8n+9n+36}=\\ =\dfrac{2n^2+17n+33}{2n^2+17n+36} \\ \text{Ramane sa demonstram ca fractia este subunitara:}\\ \text{Evident,}\ 2n^2+17n+36 >0\forall n\in \mathbb{N},\text{deci putem scapa de numitor.}\\ 2n^2+17n+33<2n^2+17n+36\\ 33<36(A)\Rightarrow \text{sirul este descrescator.}$