, 2018

Suppose on the contrary that there are infinite primes in this sequence, thus there is some index $k$ such that $a_{n}$ is composite for all $n > k$. Denote $S$ as all prime divisors of the first $k$ terms of the given sequence.

By comparing $a_{k+1}$ with the first prime $p_{1}$ that does not appear in $S$ then we have $a_{k+1} \geq p_{1}^{2}$ (since $a_{k+1}$ is composite and $p_{1}$ is not greater than the least prime divisor of this number).

Similarly, define $p_{2}$ as the second prime that does not appear in $S$ then $a_{k+2} \geq p_{2}^{2}$ (since all terms of the sequence are pairwise coprime). In general, we get
$$a_{k+i} \geq p_{i}^{2} \text{ for all } i \in \mathbb{Z}^{+} .$$

Note that $\frac{1}{\sqrt{a_{k+1} a_{k+2}}} < \frac{1}{a_{k+1}} \leq \frac{1}{p_{1}^{2}}$, $\frac{1}{\sqrt{a_{k+2} a_{k+3}}} < \frac{1}{a_{k+2}} \leq \frac{1}{p_{2}^{2}}$, $\cdots$ and so on. Then the given sum in the second condition can be divided into two parts: the first is the sum taken on values from $a_{1}, a_{2}, \ldots, a_{k}$ which is finite, and the second is less than
$$\begin{aligned} \frac{1}{p_{1}^{2}} + \frac{1}{p_{2}^{2}} + \cdots &< \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \cdots < 1 + \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots \\ &= \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) < 2 . \end{aligned}$$

Hence, the given sum is bounded, which is a contradiction and this finishes our proof.