An interesting Fibonacci discovery

Through proof by induction , we find that $\displaystyle b_n = \frac 1{2^{n-1}} + \sum_{k=1}^{n-1}\frac {L_k}{2^{n-k}}=F_n$ for all $n \ge 2$ , where $F_n$ denotes the $n$ th Fibonacci number (see Proofs below). And by induction again, we can prove that $\displaystyle a_m = a_m = \sum_{n=2}^m b_n = \sum_{n=2}^m F_n = F_{m+2}-2$ for all $m \ge 2$ . Then by numerical method, $\displaystyle \sum_{m=2}^\infty \frac 1{2+a_m} = \sum_{m=4}^\infty \frac 1{F_m} \approx 0.859829204$ . Therefore the answer is $\left \lfloor \dfrac 1{0.859829204} \right \rfloor = \boxed 1$ .

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Proof 1: $b_n = F_n$ for all $n \ge 2$

For $n=2$ , $b_2 = \dfrac 12 + \dfrac 12 = 1 = F_2$ . The claim is true for $n = 2$ . Assuming the claim is true for $n$ , then

$\begin{aligned} b_{n+1} & = \frac 1{2^n} + \sum_{k=1}^n\frac {L_k}{2^{n+1-k}} \\ & = \frac 12\left(\frac 1{2^{n-1}} + \sum_{k=1}^n\frac {L_k}{2^{n-k}}\right) \\ & = \frac 12 \left(\blue{\frac 1{2^{n-1}} + \sum_{k=1}^{n-1}\frac {L_k}{2^{n-k}}} + \frac {\red{L_n}}1\right) & \small \red{\text{Note that }L_n = F_{n-1}+F_{n+1}} \\ & = \frac {\blue{F_n} + \red{F_{n-1}+F_{n+1}}}2 \\ & = \frac {2F_{n+1}}2 = F_{n+1} \end{aligned}$

The claim is also true for $n+1$ and hence true for all $n \ge 2$ .

Proof 2: $a_m = F_{m+2} - 2$ for all $m \ge 2$

For $m=2$ , $a_2 = F_2 = 1 = F_4-2$ , the claim is true for $m=2$ . Assuming the claim is true for $m$ , then

$\begin{aligned} a_{m+2} & = \sum_{n=2}^{m+1} F_n = \sum_{n=2}^m F_n + F_{m+1} = F_{m+2} - 2 + F_{m+1} = F_{m+3} -2 \end{aligned}$

The claim is also true for $m+1$ and hence true for all $m \ge 2$ .