Trivial Fibonacci 2

Let $s_n$ denote the $n$ th summand of $a_m$ : $s_2 = \frac 12 + \frac {L_1}2 = 1,$ $s_3 = \frac 1{2^2} + \frac {L_1}{2^2} + \frac {L_2}2 = 2,$ $s_4 = \frac 1{2^3} + \frac {L_1}{2^3} + \frac {L_2}{2^2} + \frac {L_3}2 = 3,$ $s_5 = \frac 1{2^4} + \frac {L_1}{2^4} + \frac {L_2}{2^3} + \frac {L_2}{2^2} + \frac {L_3}2 = 5,$ $s_6 = \frac 1{2^5} + \frac {L_1}{2^5} + \frac {L_2}{2^4} + \frac {L_2}{2^3} + \frac {L_3}{2^2} + \frac {L_4}2 = 8.$ Evidently, $a_m$ is the sum of Fibonacci numbers. To verify this, notice that $(s_n)$ satisfies the recurrence relation $s_{n + 1} = \frac 12 s_n + \frac {L_{n - 1}}2,$ or equivalently, $2s_{n + 1} - s_n = L_{n - 1}.$ This recurrence relation is also known to be satisfied by the Fibonacci numbers: $2F_n - F_{n - 1} = L_{n - 1},$ where $F_n$ is the $n$ th Fibonacci number (with $F_0 = F_1 = 1$ ). Since $s_2 = F_1$ and $s_3 = F_2$ , and since $(s_n)$ and $(F_{n- 1})$ follow the same recurrence relation, $s_n = F_{n - 1},$ and thus $a_m = \sum_{n = 2}^m F_{n - 1} = \sum_{n = 1}^{m - 1} F_n.$ There is a simple way to compute sums of Fibonacci numbers: $\sum_{n = 1}^{m - 1} F_n = F_{m + 1} - 2.$ This can be verified by induction: when $m = 2$ , $\sum_{n = 1}^{m - 1} F_n = 1 = 3 - 2 = F_{m + 1} - 2,$ and for any $m \geq 2$ , $\sum_{n = 1}^{m - 1} F_n = F_{m + 1} - 2 \implies \sum_{n = 1}^{m} F_n = F_m + F_{m + 1} - 2 = F_{m + 2} - 2.$ This shows that $a_m = F_{m + 1} - 2$ . Now consider $\sum_{m = 2}^\infty \frac 1{(2 + a_m)^2} = \sum_{m = 2}^\infty \frac 1{(F_{m + 1})^2}.$ Since $F_{m + 1}$ increases quickly, this series converges quickly. The first twenty terms of the series give a good approximation of its value: $\sum_{m = 2}^\infty \frac 1{(F_{m + 1})^2} \approx \sum_{m = 2}^{21} \frac 1{(F_{m + 1})^2} \approx 0.176 \implies \left( \sum_{m = 2}^\infty \frac 1{(F_{m + 1})^2} \right)^{-1} \approx 5.671.$ Therefore, the solution is $\lfloor 5.671 \rfloor = \boxed 5$ .