Proof concerning Fibonacci and recursively defined sequences

Question

The series $(a_n)_{n\in\mathbb{N}}$ is given through $$a_1=1,\quad a_2=\frac{1}{2},\quad a_{n+2}=a_na_{n+1} \quad\text{ for } n\geq1.$$ I want to show that $$a_n = 2^{-f_{n-1}}$$ whereas $\,f_n$ is the $n$-th Fibonacci number.

I did a proof with induction, but now I came up with this. Out of the initial figures, the definition of $f_n$, and the exponent and logarithm rules I think it is

\begin{align} &\quad a_{n} = 2^{\,\log_2 a_n} = 2^{\,\log_2 (a_{n-2} \,\cdot\, a_{n-1})}\\ \Longleftrightarrow &\quad a_n = 2^{\,\log_2 (a_{n-2}) \,+\, \log_2 (a_{n-1})}\\ \Longleftrightarrow &\quad \log_2 a_n = \log_2 2^{\,\log_2 (a_{n-2}) \,+\, \log_2 (a_{n-1})}\\ \Longleftrightarrow &\quad \log_2 a_n = \log_2 a_{n-2} \,+\, \log_2 a_{n-1}\\ \Longleftrightarrow &\quad (-\log_2 a_n) = (-\log_2 a_{n-2}) + (-\log_2 a_{n-1}) \quad(*) \end{align}

and with the initial figures $-\log_2 a_1 = 0$ and $-\log_2 a_2 = 1$ the negative expononents form the Fibonacci sequence $2^{-f_{n-1}}$ with $(*)$.

Is that valid? Thanks!

Answer 1

I think that you should at least mention that $a_n\gt 0$ for every $n$.

Note that taking the logarithm of $a_{n+2}=a_na_{n+1}$ directly is simpler : $$\log_2 a_{n+2}=\log_2 a_n+\log_2 a_{n+1}$$

Answer 2

Your proof is fine. However, I'd suggest another version. Initialization: $a_1 = 2^{-f_0}$ (assuming $f_0 =0$), $a_2 = 2^{-f_1}$ (assuming $f_0 =1$). It happens that $a_3 = a_1*a_2 = 2^{-f_0-f_1} = 2^{-f_2}$ . Now suppose the property is valid up to index $k\ge 2$. Then $a_{k+1} = a_{k}a_{k-1} = 2^{-f_{k-1}}2^{-f_{k-2}} = 2^{-f_{k-1}-f_{k-2}} = 2^{-f_{k}}$. So it is valid for index $k+1$ as well.

Thus, you do not have to play with $\log$s, which requires positivity.