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    $\begingroup$ It doesn't work that generally. Consider $\sum_{j=1}^\infty \dfrac{n_j}{j!}$ where $n_j$ is defined recursively as the greatest integer $x$ with $\gcd(x,j!) = 1$ and $\dfrac{x}{j!} < 1 - \sum_{i=1}^{j-1} \dfrac{n_i}{i!}$. The sum of this series is a rational number, namely $1$. $\endgroup$ Commented Jul 26, 2012 at 1:59
  • $\begingroup$ (those sums should have started at $2$, not $1$) $\endgroup$ Commented Jul 26, 2012 at 2:33
  • $\begingroup$ Good point, I'll revise it. $\endgroup$ Commented Jul 26, 2012 at 3:35