$\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \cdots + \frac{1}{{{a_n}}} < 2$ 
If ${{a}_{1}},{{a}_{2}},\ldots ,{{a}_{n}}$ are distinct odd natural
   numbers not divisible by any prime greater than 5, then show that
   $\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} +  \cdots  + \frac{1}{{{a_n}}}
 < 2$.

First I have noted that $a_i$s are in the form $3^a5^b$ where $a,b$ are non-negative integers.Now let ${a_1} < {a_2} <  \cdots  < {a_n}$. As the inequality involves 2, I thought of the geometric series $1,\frac{1}{2},\frac{1}{{{2^2}}}, \cdots $ and I thought that each $\frac{1}{{{a_i}}}$ would be less than or equal to $\frac{1}{{{2^{i - 1}}}}$ and this works for some $a_i$s such as $\frac{1}{{{a_1}}} = 1 \leqslant 1, \frac{1}{{{a_2}}} = \frac{1}{3} < \frac{1}{2},\frac{1}{{{a_3}}} = \frac{1}{5} < \frac{1}{{{2^2}}}, \frac{1}{{{a_4}}} = \frac{1}{9} < \frac{1}{{{2^3}}}$. But this do not work for $i\ge 5$. I would appreciate any suggestion or idea on this problem.

$$\sum_{i=1}^n \frac{1}{a_n}<\sum_{i=0}^\infty\sum_{j=0}^\infty\frac{1}{3^i5^j}=\sum_{i=0}^\infty\frac{1}{3^i\left(1-\frac{1}{5}\right)}$$
$$=\frac{5}{4}\sum_{i=0}^{\infty}\frac{1}{3^i}=\frac{5}{4}\cdot \frac{1}{1-\frac{1}{3}}=\frac{15}{8}<2$$
So, in fact, you have a stronger statement: $\sum_{i=1}^n \frac{1}{a_n}<\frac{15}{8}$.
