# Solution to some finite sum of reciprocals of odd integers

Let it be $$S=\{O_1,O_2,...O_n\}$$ a set of an odd number of distinct odd integers, and $$O_j\notin S$$ another odd integer.

I want to prove (or disprove) that it does not exist any solution such that $$\frac{O_j-2}{O_j} = \sum_{k=1}^n{\frac{1}{O_k}}$$

Any idea?

I guess that the underlying reason for the possible inexistence of solution may be that $$O_j-2$$ and $$O_j$$ are consecutive odd integers, and that $$\gcd(O_j-2,O_j)=1$$, but I can not imagine a way to prove it or disprove it, and I do not find any counterexample.

• As the below answer shows that ths statement can be disproven (one solution is enough to do so). Maybe, the question is more interesting if we demand $O_j>3$. Commented Mar 5, 2020 at 12:45
• It really won't do to change the question after it gets an answer. If you didn't think the question through before posting it, then chalk it up to experience, accept the answer, post a new question (but leave links at each question to the other one). Anyway, the restriction seems rather ad hoc. Why the square root, and not some other function? why the sum of the $O_k$? is this coming out of some other question you want to solve, or are you making it up as you go along? Commented Mar 5, 2020 at 21:38
• @Peter, since $3/5>(1/7)+(1/9)+(1/11)+(1/13)+(1/15)+(1/17)+(1/19)$, any solution with $O_j>3$ will have $n\ge8$ and undoubtedly be hard to find. Commented Mar 5, 2020 at 21:46
• Totally agree @GerryMyerson, I will follow your advice and post a new question. Thanks! Commented Mar 5, 2020 at 22:00
• Please find it here: math.stackexchange.com/questions/3572024/… Commented Mar 7, 2020 at 6:57

$${1\over3}={1\over5}+{1\over9}+{1\over45}$$

Since someone downvoted this, maybe I have to explain.

$$O_1=5$$ is an odd integer.

$$O_2=9$$ is another odd integer.

$$O_3=45$$ is yet another odd integer.

$$3$$ is still another odd integer.

$${1\over5}+{1\over9}+{1\over45}={9\over45}+{5\over45}+{1\over45}={15\over45}={1\over3}={3-2\over3}$$

This establishes that the equation OP is asking about does, in fact, have a solution.

Capiche?

[Thanks to Jose for pointing out a typo, now corrected.]

• Thanks Gerry! Of course, the counterexample is absolutely right! Now, as suggests @Peter, lets make it a bit more interesting with some restriction (see EDIT) Commented Mar 5, 2020 at 13:03
• I think that last equation should be $$\frac{1}{3}=\frac{3-2}{3},$$ @GerryMyerson. Commented Mar 31, 2020 at 7:22