TITLE: Flaw in induction proof that the Fibonacci sequence is bounded by $(5/3)^n$
QUESTION [6 upvotes]: The Fibonacci sequence is defined by $a_1 = 1, a_2 = 1$ and for all $n \ge 2, a_{n+1} = a_n + a_{n-1}$. Thus the sequence begins
$$1,1,2,3,5,8,13,21,...$$
Prove that for all $n \ge 1, a_n < (5/3)^n$

Here is what I tried. But I am not sure what is wrong with it.
Base case, $n = 1$:
$$a_1 = 1 > \frac{5}{3}$$
Inductive step:
Assume that $a_n < (\frac{5}{3})^n$ for all $1 \le k \le n$ as the inductive hypothesis.
$$a_{n+1} = a_n + a_{n-1}$$
$$a_{n+1} < \left(\frac{5}{3}\right)^n + \left(\frac{5}{3}\right)^{n-1}$$
$$a_{n+1} < \left(\frac{5}{3}\right)^{n-1} \cdot \left(\frac{5}{3}+1\right)$$
$$a_{n+1} < \left(\frac{5}{3}\right)^{n-1} \cdot \left(\frac{8}{3}\right)$$
$$\left(\frac{8}{3}\right) < \left(\frac{5}{2}\right)^2 = \frac{25}{9}$$
$$a_{n+1} < \left(\frac{5}{3}\right)^{n-1} \cdot \left(\frac{5}{3}\right)^2$$
$$a_{n+1} < \left(\frac{5}{3}\right)^{n+1}$$

REPLY [2 votes]: A couple typos, and suggestion to consider; but substantively, there is very little "wrong" with your proof. You did the hardest part, and the proof is fine (with typos corrected!)

As pointed out in the comments, you want the direction of the inequality for the base case $a_1$ reversed. (Typo?). 
With this particular proof, I'd suggest also considering the second
base case: establishing: $$a_2 = 1 \lt \left(\frac 53\right)^2.$$
What I suspect is another typo, you typed $\left(\frac 52\right)^2$
but want $\left(\frac 53\right)^2$ in the third from the last line.
Finally, and more importantly, [here's the suggestion I am asking you
to consider]: don't be afraid to use more words in a proof: Explain
what you are doing. You need to make clear the relationships between
your lines of reasoning: Line (i) $\iff$ Line (j)? Or does one given
inequality imply the subsequent inequality? If so, indicate so, in
words or in symbols: e.g., using: "$\implies.$"