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 !
 A: Setting $b_n = 1$ (which is clearly convergent) and $a_1 = a_2 = 1$, we get
$$
a_n = \frac16(3\cdot 2^n - 3^n + 3)
$$
(WolframAlpha calculation), which doesn't converge.
A: 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.
A: This is false. Take for example $a_n=2^n$ (which is divergent) then
$$b_n=a_{n+2}-5a_{n+1}+6a_{n}=2^{n+2}-5\cdot 2^{n+1}+6\cdot 2^{n}
=(4-10+6)2^{n}=0$$
which is convergent.
A: This not true : if it were, if $b_n$ converged, you would have $c_n:=a_{n+1}-2a_n \rightarrow 0$. 
But, just setting $c_n:= 3^n + \varepsilon_n$, you have 
$$ b_n= \varepsilon_{n+1}-3\varepsilon_n  $$
you can take, then, for example, $\varepsilon_n:=\frac{1}{3^{n+1}}$, and you would have $b_n\rightarrow 0 $, $c_n\rightarrow +\infty$ and so, $(a_n)$ could not be convergent...
