Problem 1642

To solve this problem, we need to determine the probability $p$ that a randomly selected point $(x, y, z) \in \mathbb{Z}^3$ can be seen from the origin. A point $(x, y, z)$ can be seen from the origin if and only if the greatest common divisor (gcd) of $x$, $y$, and $z$ is 1, i.e., $\gcd(x, y, z) = 1$.

1. **Understanding the Probability $p$**:
The probability $p$ that a randomly selected point $(x, y, z)$ can be seen from the origin is the probability that $\gcd(x, y, z) = 1$. This is equivalent to the probability that $x$, $y$, and $z$ are coprime.

2. Using the Principle of Inclusion-Exclusion:
To find this probability, we use the principle of inclusion-exclusion. For a prime $p$, the probability that $p$ divides at least one of $x$, $y$, or $z$ is:
$$P(p \mid x \text{ or } p \mid y \text{ or } p \mid z) = 1 - P(p \nmid x \text{ and } p \nmid y \text{ and } p \nmid z)$$
Since $x$, $y$, and $z$ are independently chosen, the probability that $p$ does not divide any of them is:
$$P(p \nmid x) = P(p \nmid y) = P(p \nmid z) = 1 - \frac{1}{p}$$
Therefore,
$$P(p \nmid x \text{ and } p \nmid y \text{ and } p \nmid z) = \left(1 - \frac{1}{p}\right)^3$$
Hence,
$$P(p \mid x \text{ or } p \mid y \text{ or } p \mid z) = 1 - \left(1 - \frac{1}{p}\right)^3$$

3. Complementary Probability:
The probability that $x$, $y$, and $z$ are coprime is the product over all primes $p$ of the complementary probabilities:
$$p = \prod_{p \text{ prime}} \left(1 - P(p \mid x \text{ or } p \mid y \text{ or } p \mid z)\right) = \prod_{p \text{ prime}} \left(1 - \left(1 - \left(1 - \frac{1}{p}\right)^3\right)\right)$$
Simplifying the expression inside the product:
$$p = \prod_{p \text{ prime}} \left(1 - \left(1 - \left(1 - \frac{1}{p}\right)^3\right)\right) = \prod_{p \text{ prime}} \left(1 - \left(1 - \left(1 - \frac{3}{p} + \frac{3}{p^2} - \frac{1}{p^3}\right)\right)\right)$$
$$p = \prod_{p \text{ prime}} \left(\left(1 - \frac{1}{p}\right)^3\right)$$

4. Using the Riemann Zeta Function:
The product over all primes can be expressed using the Riemann zeta function $\zeta(s)$:
$$\prod_{p \text{ prime}} \left(1 - \frac{1}{p^s}\right) = \frac{1}{\zeta(s)}$$
Therefore,
$$p = \prod_{p \text{ prime}} \left(1 - \frac{1}{p}\right)^3 = \frac{1}{\zeta(3)}$$
Hence,
$$\frac{1}{p} = \zeta(3)$$

5. **Identifying $i$, $k$, and $s$**:
Given that $\frac{1}{p} = \sum_{n=i}^{\infty} \frac{k}{n^s}$, we recognize that this is the definition of the Riemann zeta function $\zeta(s)$ for $s = 3$, starting from $n = 1$ with $k = 1$:
$$\zeta(3) = \sum_{n=1}^{\infty} \frac{1}{n^3}$$

The final answer is $\boxed{ i = 1 }$, $k = 1$, and $s = 3$.