maybe the problem is not exatly same writen here...because the condition $ \gcd(a_i,a_j) = 1$ is extra... $ a_n - a_{n - 1} > a_{n - 1} - a_{n - 2} > .... > a_2 - a_1\geq1$ therefore $ a_n - a_{n - 1}\geq{n - 1}$...and also: $ a_k = (a_k - a_{k - 1}) + (a_{k - 1} - a_{k - 2}) + .... + (a_2 - a_1) + a_1\geq{(k - 1) + (k - 2) + .... + 1 + 1}\Rightarrow{a_k > \frac {k(k - 1)}{2}} \\ \Longrightarrow\sum^{\infty}_{i = 1}\frac {1}{a_{i}} < \frac {1}{a_1} + \sum^{\infty}_{i = 2}\frac {2}{i(i - 1)} = a_1 + 2\sum^{\infty}_{i = 2}{\frac {1}{i - 1} - \frac {1}{i}} = a_1 + 2$

Very good point, and nice solution. I have no idea what $ \gcd(a_i,a_j)=1$ is doing there.

Maybe the $ >$ should be a $ \geq$ (and the $ \gcd(a_i,a_j)=1$ still be there). Otherwise it isn't number theory at all.

ZetaX wrote: Maybe the $ >$ should be a $ \geq$ (and the $ \gcd(a_i,a_j) = 1$ still be there). Otherwise it isn't number theory at all. That looks more like a NT problem indeed. Do you have a solution for that problem?

A solution for when we only assume $a_{i+2} - a_{i+1} \ge a_{i+1} - a_i$ for all $i \ge 1$, as suggested by ZetaX almost 15 years ago: For a constant $c \in \mathbb{N}$ let $S_c = \{a_k, a_{k+1}, \ldots, a_{k+l-1}\}$ be the set of elements such that the difference $a_{k+i} - a_{k+i-1}$ is exactly equal to $c$ for all $i$ with $1 \le i \le l$. I claim that $|S_c| = l < 2c-3$ for all $c \ge 3$. Otherwise, note that $a_k, a_{k+1}, \ldots, a_{k+c-2}$ are all distinct modulo $c-1$, and so are $a_{k+c-1}, a_{k+c}, \ldots, a_{k+2c-3}$. That means that in each of these two subsequences there exists an element divisible by $c-1$, which is not allowed by the assumption $\gcd(a_i, a_j) = 1$. Likewise, $|S_1| < 3$ and $|S_2| < 5$, as otherwise there must be two integers divisible by two or three respectively. The inequalities $|S_1| < 3$ and $|S_2| < 5$ imply $a_5 - a_4 \ge 2$ and $a_{10} - a_9 \ge 3$. The inequality $|S_c| < 2c-3$ for $c \ge 3$ can now be used to show via induction, that $a_{i+1} - a_i \ge n$ holds for all $i \ge n^2$ in general. Or, stated differently, $a_{i+1} - a_i \ge \left \lfloor \sqrt{i} \right \rfloor$. This, in turn, implies that $\displaystyle a_{i+1} = a_1 + \sum_{j=1}^{i} (a_{j+1} - a_j) \ge 1 + \sum_{j=1}^i \left \lfloor \sqrt{j} \right \rfloor > \sum_{j=1}^i \frac{1}{2} \sqrt{j} > \int_0^i \frac{1}{2}\sqrt{j} \, dj = \frac{i^{3/2}}{3}$. Now the convergence of $\displaystyle \sum_{i=1}^{\infty} \frac{1}{a_i}$ is clear.