<s>[INST] Proof by induction: $a_n > n + \frac{1}{3} \space \forall n \in N,\space n\geq 4$ How do I prove this by induction?
Let $(a_n)_{n\in N}$ be the sequence defined by:
$$a_1=1,\space a_2=\frac{3}{2},\space a_{n+2}=a_{n+1}+\frac{2n+1}{n+2}a_n \space \space (n\in N)$$ 
Prove that $a_n > n + \frac{1}{3} \space \forall n \in N,\space n\geq 4$
Here's what I did, is it ok?
1) I define

$$P(n): a_n > n + \frac{1}{3} \space \forall n \in N,\space n\geq4\\$$

2) Since the base case is true,
$$a_1=1, \ a_2=\frac{3}{2},\ a_3=\frac{5}{2},\ a_4=5 $$
$$P(4):a_4 = 5 > 4+\frac{1}{3}$$
3) I assume $P(n)$ (already defined) and

$$P(n+1):a_{n+1}>n+1+\frac{1}{3} = n + \frac{4}{3}$$

are true, and want to prove that it implies

$$P(n+2):a_{n+2}>n+2+\frac{1}{3}=n+\frac{7}{3}$$

is true.
4)So since:
by the second induction hypothesis,
$$a_{n+2}=a_{n+1}+\frac{2n+1}{n+2}a_n > n+\frac{4}{3}+\frac{2n+1}{n+2}a_n\ $$ 
and by the first induction hypothesis,
$$n+\frac{4}{3}+\frac{2n+1}{n+2}a_n > n+\frac{4}{3}+\frac{2n+1}{n+2}(n + \frac{1}{3}) = n+\frac{4}{3} + \dfrac{2n^2+n+\dfrac{2n}{3}+\dfrac{1}{3}}{n+2} $$
5)Now as I know that every term is positive I can remove some terms and assure that:
$$n+\frac{4}{3} + \dfrac{2n^2+n+\dfrac{2n}{3}+\dfrac{1}{3}}{n+2} > n+\frac{4}{3}+\frac{2n+1}{n+2}(n + \frac{1}{3}) = n+\frac{4}{3} + \dfrac{2n^2}{n+2}$$
So the only step remaining would be making sure that:
$$\dfrac{4}{3}+\dfrac{2n^2}{n+2} > \dfrac{7}{3}\ $$
then
$$ \dfrac{2n^2}{n+2} > 1 which \ is \ true \ \forall n \in N, n \geq 4$$
Therefore $P(n+2)$ is true, $P(n+1)$ is true and also $P(n)$ is true.
 [/INST] What you did is correct. A shortcut would be to notice that 
$$a_{n+2}=a_{n+1}+\frac{2n+1}{n+2}a_n > n+\frac{4}{3}+ \underbrace{\frac{2n+1}{n+2}a_n}_{\geq 1} > n+\frac{4}{3}+1 = n + \frac{7}{3}$$
since we have that $a_n > 4 + \frac{1}{3} > 1$ and $\frac{2n+1}{n+2} > 1$ are true for all $n \geq 4$. Hence, $\frac{2n+1}{n+2}a_n >1$ for all $n \geq 4$.
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