[tex]\text{Cel mai simplu ar fi sa aratam ca }\dfrac{a_{n+1}}{a_n}<1\\
\dfrac{a_{n+1}}{a_n}=\dfrac{\frac{n+2}{n+1}}{\frac{n+1}{n}}=\dfrac{n+2}{n+1}\cdot \dfrac{n}{n+1}=\dfrac{(n+2)n}{(n+1)^2}=\dfrac{n^2+2n}{n^2+2n+1}\\
\text{Mai departe avem ca:}\\
\dfrac{n^2+2n}{n^2+2n+1}<1\\
n^2+2n<n^2+2n+1\\
0<1(A)\Rightarrow \text{sirul este descrescator}[/tex]