Question

Sa se studieze monotonia sirului :

Answer 1

Salut,

$a_n=\dfrac{n(n+2)}{n+1}=\dfrac{n(n+1+1)}{n+1}=\dfrac{n(n+1)}{n+1}+\dfrac{n}{n+1}=n+\dfrac{n}{n+1}.\\\\Deci\ a_n=n+\dfrac{n}{n+1}\Rightarrow a_{n+1}=n+1+\dfrac{n+1}{n+2}.\\\\a_{n+1}-a_n=1+\dfrac{n+1}{n+2}-\dfrac{n}{n+1}=1+\dfrac{(n+1)^2-n^2-2n}{(n+1)(n+2)}=\\\\=1+\dfrac{1}{(n+1)(n+2)}>0.\\\\Cum\ a_{n+1}>a_n,\ rezult\breve{a}\ c\breve{a}\ a_n\;este\;strict\;cresc\breve{a}tor.$