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Conjecture:

If $c_n$ is the n'th composite number,$$\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\cdots+\frac{1}{c_n}\not\in \Bbb{N}$$ for any $n$ value.

Inspiration: I was originally intrigued by the idea of how this never works for natural numbers, and so I then thought of this question, which I have been pondering ever since. I feel, as the probability of like factors is greater, that this has a greater chance of summing to a natural number. However, the more I have thought about it, the more complex it becomes.

Attempts: I have really not understood where to start mathematically, but have used very loose logical intuition to take a jab here. I managed to concur that if there are no solutions for low $n$ values, then there is a lower chance for a solution for upper $n$ values because there is a larger chance that the composite denominator have factors of extremely obscure primes (like 17 or 31, you get the idea).

As an answer, please prove or disprove my conjecture. A proof would be...well a proof, and a disproof might be a $n$ value that adds up to a natural number (so state the $n$). Thanks in advance. I think I am really liking it here on Stack Exchange.

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As, André Nicolas's comment, that will be the best way to approach.

Let $2^k$ be the largest power of $2$ less than or equal to $c_n$. Now, consider $$\color{red}{l=lcm(2,4,6,8,9,\dots, c_n)},$$ clearly, $\color{blue}{l}$ is composite.

Now, consider the sum $$\color{red}{S=\frac l2+\frac l4+\frac l6+\frac l8+\frac l9+\dots+\dots+\frac l{2^k}+\dots+\frac l{c_n}}.$$

Look that, the fraction $\color{blue}{\frac Sl}$ equals your required sum.

Now, see that $\color{blue}{S}$ is odd, since, all the even factors of $\color{blue}{l}$ i.e. $2^k$ vanishes at the one only term where in the denominator, $\color{blue}{2^k}$ comes. And all the else terms are composite, because there all the even factors of $\color{blue}{l}$ does not vanish.

So, ultimately, $\color{blue}{S}$ is odd and $\color{blue}{l}$ is even, so the frac $\color{blue}{\frac Sl}$ never reduces to an integer.

Hence your claim.

Note: The prime counterpart of this question is answered here.

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  • $\begingroup$ And this is just the same as the usual elementary proof that $\frac11+\frac12+\frac13+\frac14+\cdots+\frac1n$ is a non-integer for $n\ge2.$ $\endgroup$ – bof Nov 2 '15 at 7:23
  • $\begingroup$ @bof,Yes, of course. $\endgroup$ – user249332 Nov 2 '15 at 7:49
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It would appear that there could be a nifty proof of this adapted from the very same proof that there is only one harmonic number that is also an integer. See the following link:

https://math.stackexchange.com/a/2752/286106

Basically, go for contradiction. Assume you can stop somewhere and get an integer, and call it $m$. Find the denominator with the highest power of $2$, say $2^p$, and multiply the whole equation through by $2^{p-1}$. You'll get something like this: $$ m = \frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots+\frac{1}{c_k} $$ $$ 2^{p-1}m=2^{p-3}+\frac{n_2}{d_2}+\frac{n_3}{d_3}+\cdots+\frac{1}{2}+\cdots+\frac{n_k}{d_k} $$

Now, I've labelled the indices to make them match up with their corresponding composite number in the sum; since the first number is $\frac{1}{4}$, it's denominator will be $d_1=1$ as long as $p\geq3$, our friend $\frac{1}{2^p}$ turns into $\frac{1}{2}$, and everyone else is stuck with an odd denominator! Now rearrange to find: $$ -\frac{1}{2} = 2^{p-3}+\frac{n_2}{d_2}+\frac{n_3}{d_3}+\cdots-2^{p-1}m+\cdots+\frac{n_k}{d_k} $$

But lo! the sum on the right, after reducing, must have an odd denominator, while the number on the left has an even denominator! This contradiction tells us that $m$ could not have existed in the first place unless $p=2$. Now simply check the finitely remaining cases, and we are done -- or rather, assuming I haven't made any mistakes, we have proven your conjecture.

P.S. Looks like I got beaten to the punch!

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