Problem 1453

1. **Prove for $n \ge 4$:**
We start by considering the sum of the reciprocals of the squares of the first $n$ prime numbers:
$$\frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2} + \cdots + \frac{1}{p_n^2}$$

2. Use the infinite series bound:
We know that the sum of the reciprocals of the squares of all prime numbers is bounded by the sum of the reciprocals of the squares of all natural numbers:
$$\sum_{k=1}^{\infty} \frac{1}{p_k^2} \le \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$$
Since the partial sums form a monotone increasing sequence, each partial sum is less than the infinite sum:
$$\sum_{k=1}^{n} \frac{1}{p_k^2} \le \sum_{k=1}^{\infty} \frac{1}{p_k^2}$$

3. Find a tighter upper bound:
We subtract the contributions of 1 and all even numbers greater than 2 from the sum of the reciprocals of the squares of all natural numbers:
$$\sum_{k=1}^{\infty} \frac{1}{p_k^2} \le \left( \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots \right) \le \frac{\pi^2}{6} - 1 - \left( \frac{1}{4^2} + \frac{1}{6^2} + \cdots \right)$$
The sum of the reciprocals of the squares of all even numbers greater than 2 is:
$$\sum_{k=2}^{\infty} \frac{1}{(2k)^2} = \frac{1}{4} \sum_{k=2}^{\infty} \frac{1}{k^2} = \frac{1}{4} \left( \frac{\pi^2}{6} - 1 \right)$$
Therefore:
$$\sum_{k=1}^{\infty} \frac{1}{p_k^2} \le \frac{\pi^2}{6} - 1 - \frac{1}{4} \left( \frac{\pi^2}{6} - 1 \right) = \frac{3}{4} \left( \frac{\pi^2}{6} - 1 \right) \approx 0.4837 < \frac{1}{2}$$

4. Add the extra term:
We now add the term $\frac{1}{p_1 p_2 \cdots p_n}$. For $n \ge 4$, the product of the first $n$ primes is large, making this term very small. For example:
$$\frac{1}{p_1 p_2 p_3 p_4} \approx 0.00476$$
Since $\frac{1}{p_1 p_2 \cdots p_n}$ decreases as $n$ increases, we have:
$$\frac{1}{p_1 p_2 \cdots p_n} < \frac{1}{p_1 p_2 p_3 p_4}$$

5. Combine the results:
Adding the largest terms gives us the upper bound for all the partial sums of inverse squares plus the extra product term:
$$\frac{1}{p_1^2} + \frac{1}{p_2^2} + \cdots + \frac{1}{p_n^2} + \frac{1}{p_1 p_2 \cdots p_n} < \sum_{k=1}^{\infty} \frac{1}{p_k^2} + \frac{1}{p_1 p_2 p_3 p_4} \le \frac{3}{4} \left( \frac{\pi^2}{6} - 1 \right) + 0.00476 \approx 0.4885 < \frac{1}{2}$$

6. **Special case $n = 3$:**
For $n = 3$, we calculate directly:
$$\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{2 \cdot 3 \cdot 5} \approx 0.25 + 0.1111 + 0.04 + 0.0333 \approx 0.4344 < \frac{1}{2}$$

Hence, we have our result.