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?

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.


  • 2
    $\begingroup$ If $p>3$ is prime, then $p$ divides $\frac{10^{p-1}-1}{9} = \underbrace{11 \dots 1}_{p-1}$, so in fact each prime except $2$ occurs as a prime factor of one of the numbers $11$, $111$, $1111$, ... $\endgroup$
    – user133281
    Apr 25, 2016 at 15:25
  • $\begingroup$ @user133281: Thanks for answering the implicit question within my post (was wondering about that too). $\endgroup$ Apr 25, 2016 at 15:28

1 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)

  • 2
    $\begingroup$ This is a very nice counterexample. Thanks!!! $\endgroup$ Apr 25, 2016 at 9:17

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