Problem 1461

1. We start by considering the sequence $a_n = \frac{1}{F_n F_{n+2}}$ for all $n \ge 0$. We need to prove that the sum of the first $m+1$ terms of this sequence is less than 1 for all $m \ge 0$.

2. First, we use the identity for Fibonacci numbers:
$$F_{n+2} = F_n + F_{n+1}$$
This allows us to rewrite $a_n$ as follows:
$$a_n = \frac{1}{F_n F_{n+2}} = \frac{1}{F_n (F_n + F_{n+1})}$$

3. We can further manipulate this expression:
$$\frac{1}{F_n (F_n + F_{n+1})} = \frac{1}{F_n F_{n+2}} = \frac{F_{n+2} - F_n}{F_{n+1} F_n F_{n+2}}$$
Simplifying, we get:
$$\frac{1}{F_n F_{n+2}} = \frac{1}{F_{n+1}} \left( \frac{1}{F_n} - \frac{1}{F_{n+2}} \right)$$

4. Now, we sum the terms from $n = 0$ to $n = m$:
$$\sum_{i=0}^m \frac{1}{F_i F_{i+2}} = \sum_{i=0}^m \frac{1}{F_{i+1}} \left( \frac{1}{F_i} - \frac{1}{F_{i+2}} \right)$$

5. Notice that this is a telescoping series. The terms inside the summation will cancel out in pairs:
$$\sum_{i=0}^m \frac{1}{F_{i+1}} \left( \frac{1}{F_i} - \frac{1}{F_{i+2}} \right) = \left( \frac{1}{F_0 F_1} - \frac{1}{F_{m+1} F_{m+2}} \right)$$

6. Given that $F_0 = 1$ and $F_1 = 1$, we have:
$$\frac{1}{F_0 F_1} = 1$$

7. Therefore, the sum becomes:
$$\sum_{i=0}^m \frac{1}{F_i F_{i+2}} = 1 - \frac{1}{F_{m+1} F_{m+2}}$$

8. Since $F_{m+1}$ and $F_{m+2}$ are both positive integers and grow exponentially, $\frac{1}{F_{m+1} F_{m+2}}$ is a positive term that decreases as $m$ increases. Thus:
$$1 - \frac{1}{F_{m+1} F_{m+2}} < 1$$

9. Hence, we have shown that:
$$\sum_{i=0}^m a_i < 1$$
for all $m \ge 0$.

$\blacksquare$