I came across the following recreational problem and am not sure if I did it right:
Let $x_1, \ldots, x_n$ be odd numbers with a prime divisor not greater than 5. Prove that it must hold that $$\sum\limits_{i=1}^{n} \frac{1}{x_i} < \frac{15}{8}$$
How would you tackle the problem?
I thought first that, since all $x_i$ are odd it is $x_i \geq 3$. But now I am not sure I understand the problem correct since if $n$ is not specified I could choose $n$ large enough such that $n\frac{1}{3} > \frac{15}{8}$ in case it is $x_1 = \ldots = x_n = 3$ because the exercise does not restrict to distinct $x_i$. In case all $x_i$ are distinct (and $n$ is finite) we could just set up the inequality
$$\sum\limits_{i=1}^{n} \frac{1}{x_i} < \lim\limits_{n\to \infty} \sum\limits_{i=0}^{n}\sum\limits_{j=0}^{n}\frac{1}{3^i5^j} = \frac{15}{8}$$
but the exercise presumebly also allows for negative numbers as it does not specifically mentions positive odd numbers, so I am not sure about my result.
Since $\mathbb{Z}$ is a unique factorization domain
$$ \sum_{b,c\geq 0}\frac{1}{3^b 5^c}=\left(\sum_{b\geq 0}\frac{1}{3^b}\right)\left(\sum_{c\geq 0}\frac{1}{5^c}\right) = \frac{3}{2}\cdot\frac{5}{4} = \frac{15}{8} \tag{1}$$ and the LHS of $(1)$ is exactly $$ \sum_{\substack{n\text{ odd}:\\p\mid n \Rightarrow p\leq 5}}\frac{1}{n}.\tag{2} $$