If $b_n$ is convergent, then $a_n$ is also convergent 
Let $(a_n)_{n\geq 1}$ and $(b_n)_{n \geq 1}$ be two sequences of real numbers such that $$b_n=a_{n+2}-5a_{n+1}+6a_{n}, \: \forall n \geq 1$$
  Prove that if $(b_n)$ is convergent, then $(a_n)$ is also convergent.

I defined $c_n=a_{n+1}-2a_n$ and the relation became $b_n=c_{n+1}-3c_n.$ Then I tried to prove that $c_n$ is convergent by expressing $c_n$ only in terms of $b_n, b_{n-1}, \dots b_1$ and $c_1$, but the convergence doesn't follow from here and I got stuck.
EDIT: As proven below, this statement is false !

Define $c_n=a_{n+1}-a_n$, then the recursion transforms to:
$$b_n=c_{n+1}-6c_n$$ 
If $c_n$ converges then $b_n$ converges. So, to disprove the statement it suffices to find a non-convergent sequence $a_n$ such that $a_{n+1}-a_n$ converges. 
This is a classical problem. One such example is $a_{n}=\sqrt{n}$. 
One can also find examples where $a_n$ is bounded and divergent.
