Bounded sequence metric spaces - real analysis

Question

Show that there is a unique bounded sequence $\{a_n\}$$_{n\in \mathbb N}\subset \mathbb R$ such that

$$\forall n\in \mathbb N \ \ \ a_{n+1} + a_{n+2} = (n + 2)a_n + 1$$ I am unsure as to how to approach this question. Any help would be great.

Answer 1

It is not too difficult to show that there is at most one bounded sequence as above: suppose for a contradiction that there are two different bounded sequences $a_n$ and $b_n$. Then the sequence $c_n:=a_n-b_n$ obviously satisfy the relation $$c_{n+2}=(n+2)c_n-c_{n+1}.$$

Moreover, notice that by the hypothesis that the sequences are different we can conclude that there is no $m$ such that $c_m=c_{m+1}=0$, since otherwise $c_n$ would be the constant $0$ sequence. So we deduce that there are infinitely many $n$'s such that $c_n\neq 0$. We assume that for infinitely many $n$ $c_n>0$ (this can be done, maybe at the cost of considering the sequence $-c_n$, which clearly satisfies the same relation as $c_n$).

Let $B:=\sup\{c_n: n\geq 0\}$. It follows that $B>0$. But also, for every $\epsilon$, there is an $n$ such that $B-c_n<\epsilon$. Suppose $n>0$, then we have that $$B\geq c_{n+2}=(n+2)c_n-c_{n+1}\geq (n+2)(B-\epsilon)-B,$$ which gives $B\leq \epsilon(n+2)/n$, which implies that $B=0$, a contradiction. This implies that $c_0=B$, but then $c_2=2B-c_1$ forces $c_1=c_2=B$, which gives rise to an unbounded sequence, and this is a contradiction since the difference of two bounded sequences is bounded.

Anyway, I wasn't able to prove that there is a sequence satisfying the conditions.