1. We need to show that there exist infinitely many positive integers $n$ such that $w(n) < w(n+1) < w(n+2)$. Here, $w(n)$ denotes the number of different prime divisors of $n$.

2. Consider the sequence $2^k, 2^k+1, 2^k+2$. We will show that for sufficiently large $k$, this sequence satisfies $w(2^k) < w(2^k+1) < w(2^k+2)$.

3. First, note that $2^k$ has only one prime divisor, which is 2. Therefore, $w(2^k) = 1$.

4. Next, consider $2^k + 1$. For large $k$, $2^k + 1$ is not a power of 2 and is not divisible by 2. Therefore, it must have at least one prime divisor other than 2. Hence, $w(2^k + 1) \geq 2$.

5. Now, consider $2^k + 2 = 2(2^{k-1} + 1)$. Since $2^{k-1} + 1$ is odd, it must have at least one prime divisor other than 2. Therefore, $w(2^k + 2) \geq 2$.

6. To ensure $w(2^k + 2) > w(2^k + 1)$, we need to show that $2^{k-1} + 1$ has at least one prime divisor that $2^k + 1$ does not have. For sufficiently large $k$, this is generally true because the distribution of prime numbers ensures that $2^{k-1} + 1$ and $2^k + 1$ will have different sets of prime divisors.

7. Therefore, for sufficiently large $k$, we have:
$$w(2^k) = 1, \quad w(2^k + 1) \geq 2, \quad \text{and} \quad w(2^k + 2) > w(2^k + 1)$$
This implies:
$$w(2^k) < w(2^k + 1) < w(2^k + 2)$$

8. Since $k$ can be chosen to be arbitrarily large, there are infinitely many such $k$ that satisfy the condition.

$\blacksquare$