Is this a valid argument for proving that a sum of reciprocals is irrational?

Question

Suppose we have a strictly increasing sequence of natural numbers.

Suppose that the sum of the reciprocals of the elements converges.

And suppose that the elements have infinitely many prime factors.

Does this imply that the sum of the reciprocals of the elements is irrational?

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My only thought is - yes:

• We can never find a common denominator for all elements
• So there is no way to represent the sum as a simple fraction

My motivation comes from this question on $\frac{1}{11}+\frac{1}{111}+\frac{1}{1111}+\dots$

I'm not even sure that $11,111,1111,\dots$ have infinitely many prime factors.

But I was still wondering if this argument (assuming it's correct) could be used.

Thanks

Answer 1

In the general case, this argument is false.

You can use the Sylvester sequence, defined this way : $a_0 = 2$ and $a_{n+1} = \prod \limits_{k=0}^n a_k + 1$.

Hence $a_0 = 2$, $a_1 = 3$, $a_2 = 7$, $a_3 = 43$, $a_4 = 1807$...

You can prove that $a_{n+1} = a_n(a_n - 1) + 1$, and also by induction that $\sum \limits_{k=0}^n \frac{1}{a_k} = 1 - \frac{1}{a_{n+1} - 1}$.

So $\sum \limits_{k=0}^{+\infty} \frac{1}{a_k} = 1$, and the $a_k$ are all coprime (and $1$ is rational)