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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.

Taufi
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  • Obviously the numbers must be distinct (as you have seen by yourself). Also negative numbers only can make the sum smaller. So what is the problem exactly? May it be that the question is not phrased very formally and that "numbers" actually mean "distinct positive natural numbers"? In this sense I think your result is correct and complete. – M. Winter Jun 18 '17 at 16:17
  • Okay, I just wanted to know whether I missed something here since the question is not phrased precisely as you already mentioned. – Taufi Jun 18 '17 at 16:23

1 Answers1

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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 $$ \sum_{\substack{n\text{ odd}:\\p\mid n \Rightarrow p\leq 5}}\frac{1}{n}.\tag{2} $$

Jack D'Aurizio
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