# Another criterion for irrationality of real numbers?

here's a link to a very elegant proof (and the simpler that I know of) that $\mathrm{e}$ is irrational: https://math.stackexchange.com/posts/713472/edit. What it boils down to is that ${n!}\mathrm{e}$ cannot be an integer for any natural number $n$ which is convincing. The proof is based on the very fast convergence of the series for $\mathrm{e}$, so I thought if it where possible to extend this for other numbers that are given as an infinite series. So suppose we are given a number as a limit of a series: $${S}_{n}=\sum_{k=1}^n {a_k}, \space a_k>0, \space S=\lim_{n \to \infty} S_n$$ My idea is to show that ${c_n}{S_n}$ cannot be an integer. If we can find two sequences $b_k>0$ and $c_k>0$ such that $$b_k<1, \space 0<S-S_n<\frac{b_n}{c_n}, \space \text{and} \space {c_n}{a_k} \space \text{is an integer for} \space k=1,2,...n$$ then we will have $$0<c_nS-c_nS_n<b_n \space \text{or} \space c_nS_n<c_nS<c_nS_n+b_n.$$ Now noting that $$c_nS_n=\sum_{k=1}^nc_na_k$$ is an integer and $b_n<1$ completes the proof that $c_nS_n$ cannot be an integer.

In order for this to work now the $c_n$ sequence has to visit all natural numbers at least once. It could be $n$, $n!$ or even $\sqrt{n}$ for example.

Question 1: Can we think of any examples other than $e$ that this could work on?

Question 2: Is it really so simple? Does it only depend on the rate of convergence? If we know of a series that converges to an irrational number is it true that any other series with faster rate of convergence will also converge to an irrational limit? I would be surprised if that holds. It would be amazing! Perhaps we can come up with some counter-example.

Just food for thought...

• Unfortunately, such a proof does not work for $\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\cdots$. – Peter Apr 19 '17 at 11:09