Let $a_1 , a_2 > 0$ and for $j \ge 3$ define $a_j = a_{j-1} + a_{j-2}$. Show that this sequence cannot converge to a finite limit. I was trying this problem and I wonder if I could have some feedback for my solution.

Let $a_1 , a_2 > 0$ and for $j \ge 3$ define $a_j = a_{j-1} + a_{j-2}$. Show that this sequence cannot converge to a finite limit.

Suppose $\{a_j\}$ converges to finite number $L$.
Then, let $\epsilon = a_1 > 0$, and there exists $N$ such that $|a_j-a_{j+1}| < \epsilon = a_1$ for all $n > N$.
$$|a_j-a_{j+1}| = |a_j+a_{j-1}-a_j| = |a_{j-1}|$$
However, since $|a_{j-1}| > \epsilon = a_1$ for all $n$, it contradicts, and the sequence diverges.
Is this right? or is there anybody who can give some hints?
Thank you.
 A: Both places where you’ve written $n$, you meant $j$. Also, we can only guarantee that $a_j>a_1$ for $j\ge 3$, since $a_2$ could be less than or equal to $a_1$. Those are small technical errors, and apart from them the argument is correct, except that you really ought to include justification for the fact that $a_j>a_1$ for $j\ge 3$, even though it’s a very easy proof by induction.
You might be interested in an alternative approach. Let $a=\min\{a_1,a_2\}$; clearly $a_2\ge a$ and $a_3\ge 2a$. Suppose that $n>3$, and $a_j\ge(j-1)a$ for $2\le j<n$. Then
$$a_n=a_{n-1}+a_{n-2}\ge(n-2)a+(n-3)a\ge(n-1)a\,,$$
and by induction $a_n>(n-1)a$ for all $n\ge 2$. The sequence is therefore unbounded and cannot converge to a finite limit.
If you’re familiar with the Fibonacci numbers, you can use this same approach to show that $a_n\ge F_na$ for $n\ge 1$. It is known that $F_n$ is the integer closest to $\frac{\varphi^n}{\sqrt5}$, where $\varphi=\frac12\left(1+\sqrt5\right)\approx 1.618>1$, so in fact the sequence $\langle a_n:n\ge 1\rangle$ grows exponentially fast.
A: Another Idea : let $$\min \{ a_1 , a_2\}=m$$ and we know $m>0 $ so
$$a_3=a_2+a_1\geq m+m=2m$$
and
$$a_4=a_3+a_2\geq2m+m=3m\\a_5=a_4+a_3 \geq 3m+2m >4m\\\vdots\\a_n\geq(n-1)m ,m >0 \\\to \lim_{n\to \infty}a_n\to +\infty$$
