Convergence of Series Whose Terms are Defined Recursively My recursively defined sequence $(a_n)_{n\in\mathbb{N}}$ is given trough

$$a_1 = 1, \quad a_2=\frac{1}{2}\quad a_{n+2}=a_{n}a_{n+1}\quad \text{for } n\geq1$$

and I have to show that the series

$$\sum_{n=1}^{\infty}a_n$$

converges.



I have 4 questions to my approach(es) I did to deal with that:


I noticed that $a_n$ can be represented as $$a_n = 2^{-f_{n-1}}$$
whereas $f_n$ is the $n$-th Fibonacci number. For this claim I did this proof by induction:

$$ n=1: a_1 = 1 = 2^0 = 2^{f_0} = 2^{-f_0} = 2^{-f_{1-1}} \\
 n=2: a_2 = \frac{1}{2} = 2^{-1} = 2^{-f_1} =  2^{-f_{2-1}}$$
and by multiplying the premises $A(n)$ and $A(n+1)$ in order to get $A(n+2)$
$$"n+1\to n+2":\quad a_n \cdot a_{n+1}  = 2^{-f_{n-1}} \cdot 2^{-f_n} \quad\Longleftrightarrow\\  a_{n+2} = a_n \cdot a_{n+1} = 2^{-f_{n-1}} \cdot 2^{-f_n} = 2^{-(f_n + f_{n-1})} = 2^{-f_{n+1}}.$$
...is that valid?
Given that $a_n = 2^{-f_{n-1}}$ is true I proceeded with the ratio test
$$q := \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \left|\frac{2^{-f_{n}}}{2^{-f_{n-1}}}\right| = \lim_{n\to\infty} \frac{2^{f_{n-1}}}{2^{f_{n}}} =\\
 = \lim_{n\to\infty} \frac{1}{2^{f_{n} - f_{n-1}}} = \lim_{n\to\infty}\frac{1}{2^{(f_{n-2} + f_{n-1}) - f_{n-1}}} = \lim_{n\to\infty}\frac{1}{2^{f_{n-2}}} &lt; 1.$$
Is that sufficient to show the convergence? Follow-up question: given the $a_n = 2^{-f_{n-1}}$ term - is there a better way than this ratio test to show convergence of $\sum a_n$ now that I have this representation?
Another approach would be to find a sequence $(b_n)_{n\in\mathbb{N}}$ with
$$\sum_{n=1}^{\infty}b_n &lt; \infty \quad\wedge\quad |b_n| &gt; |a_n|\quad\forall n\geq1.$$
For what it's worth I struggle to find such a series - am I missing something? Is there something immediate that should come to mind? $a_n$ and $b_n$ have to be zero sequences, how can I do the estimate in its recursively defined form?
Another "please correct me if I'm wrong"-question: I did an approach where I proved that $a_{n+1}&lt; a_{n}$ for all $n\geq1$ (which is similar to question 1:

$$n=1:\quad a_1 = 1 \geq \frac{1}{2} = a_2\\
n=2: a_3 = a_1a_2 = 1 \cdot \frac{1}{2} = \frac{1}{2},\quad\therefore a_2 = \frac{1}{2} \geq \frac{1}{2} = a_3$$
and
$$"n+1 \to n+2":\quad a_n \cdot a_{n+1} \geq a_{n+1} \cdot a_{n+2} \Longleftrightarrow\\
a_{n+2} = a_n \cdot a_{n+1} \geq a_{n+1} \cdot a_{n+2} =  a_{(n+1)+2} = a_{n+3}$$
respectively) and after that I did the ratio test
$$q := \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \left|\frac{a_{n+2}}{a_{n+1}}\right| = \lim_{n\to\infty} \left|\frac{a_n a_{n+1}}{a_{n+1}}\right| = \lim_{n\to\infty}\left|a_{n}\right| \stackrel{*}{&lt;} 1.$$
(with (*) denoting that $a_{n+1}&lt;a_{n}$ and $a_2 &lt;1$). Is that a sufficient answer to my problem?




...thank you so much, guys!
I would greatly prefer your approach no.3 because such series are actually easy to find. One example is the geometric series with ratio 1/2 (up to an initial "stutter" term which is easy to discard).
I would greatly prefer your approach no.3 because such series are actually easy to find. One example is the geometric series with ratio 1/2 (up to an initial "stutter" term which is easy to discard).