I need to prove that f(n) = Ѳ(n)

when f:N -> R defined by

$\displaystyle \frac{n^2+2log2n}{n+1}$

<= since log2n<=n

(I left out these works)

= 6n for n>0

since each term >= 0

>= since log2n is >= 0 for n>=1

>=$\displaystyle \frac{n^2}{n+n}$

= $\displaystyle \frac{n}{2}$ for n>0

Therefore f(n) = Ѳ(n)

is it ok?