convergence of a series involving reciprocals of increasing sequence of positive integers A known problem (published in many books and teaching resources):
Let $(x_n)$ be an increasing sequence of positive integers such that $\lim(x_{n+1}-x_n)=\infty$.
Is $\sum\frac{1}{x_n}$ convergent?
It has negative answer, the standard counterexample involves the floor of a logarithm.
I have added the condition that $x_n$'s are relatively prime, and the problem turned out to be too difficult for me, that's why I aks it here:
Let $(x_n)$ be an increasing sequence of positive integers such that $\lim(x_{n+1}-x_n)=\infty$ and $\gcd(x_i,x_j)=1$ for $i\neq j$.
Is $\sum\frac{1}{x_n}$ convergent?
 A: Consider the sequence $(x_n)$ such that $x_1 = 3$ and for $n \geq 1$, $x_{n+1}$ is the smallest prime greater than $x_n$ with $x_{n+1} - x_n \geq \log \log x_n$. Then with this definition, the $x_n$'s are distinct primes, hence relatively prime, and we have $\lim \,(x_{n+1} - x_n) \geq \lim \log \log x_n = \infty$. 
Now, let $s(k)$ be the number of elements $x_n$ of the sequence with $2^{k-1} < x_n \leq 2^k$. Note the number of primes in this interval is $\pi(2^k) - \pi(2^{k-1}) \sim \frac{2^{k-1}}{k \log 2}$ by the prime number theorem, where $\pi(x)$ is the prime counting function. The idea is that at least roughly every $(\log k)$-th prime in the interval is a member of our sequence. To make this precise, note that if $x_n = p_m \leq 2^k$ and $x_{n+1} = p_{m+r}$, then $r \leq \lceil \log \log x_n \rceil < \log k + 1$, so the number of primes between $x_n$ and $x_{n+1}$ is at most $\log k$. Thus letting $x_t, x_{t+1}, \dots, x_{t+s(k)-1}$ be the members of the sequence in $(2^{k-1}, 2^k]$, there are at most $\log k$ primes between each $x_i$ and $x_{i+1}$, as well as at most $\log k$ between $2^{k-1}$ and $x_t$ and at most $\log k$ between $x_{t+s(k)-1}$ and $2^k$, giving a total of at most $(s(k) + 1) (\log k + 1)$ primes in the interval. This means $s(k) \geq \frac{\pi(2^k) - \pi(2^{k-1})}{\log k + 1} - 1$, so for large $k$ we have $s(k) \geq c \frac{2^k}{k \log k}$, hence $\frac{s(k)}{2^k} \geq \frac{c}{k \log k}$. But the series $\sum \frac{1}{k \log k}$ diverges, so our series $\sum \frac{1}{x_n} \geq \sum \frac{s(k)}{2^k}$ diverges as well.
