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I recently saw a pretty elegant proof of the irrationality of $e$, namely:

Let $s_n:=\sum_{k=0}^{n}{\frac{1}{k!}}$ such that $e=\lim_{n\to\infty} s_n$. We obviously have $s_n<e$ and furthermore $e=s_n+\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+...=s_n+\frac{1}{n!}\left(\frac{1}{(n+1)}+\frac{1}{(n+1)(n+2)}+...\right)<s_n+\frac{1}{n!}\left(\frac{1}{(n+1)}+\frac{1}{(n+1)^2}+...\right)=s_n+\frac{1}{n!n}$. Therefore, we have $n!s_n<n!e<n!s_n+\frac{1}{n}$. If $e$ was rational, then $e=\frac p q$ for $p,q\in\mathbb N$. But since we can choose $n$ arbitrarily high we can assume $n>q$. Then the numbers $n!s_n$ and $n!e$ are clearly integers and we get $n!s_n<n!e<n!s_n+\frac{1}{n}≤n!s_n+1$, contradiction.

I adapted this proof to show the irrationality of numbers like $\sum_{k=0}^{\infty} \frac{1}{2^{n^2}}$ which for me raised the following question: This kind of proof seems to be working well with rapidly converging series. Can there be said anything rigorous about the irrationality of such series in order to justify this observation?

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I would say that $\sum_0^\infty \frac{1}{n!}$ is irrational for a "different" reason than $\sum_0^\infty \frac{1}{2^{n^2}}$, and rapidity of convergence is not the full reason for the irrationality of either sum.

Let $a_0,a_1,a_2,a_3,\dots$ be a sequence of rationals that is very rapidly converging to $0$. Let $b_i=a_{i}-a_{i+1}$. Then the sequence $b_0,b_1,b_2, b_3,\dots$ converges very rapidly to $0$, and yet $\sum_{i=0}^\infty b_i$ is, by telescoping, equal to the rational $a_0$.

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  • $\begingroup$ Thanks! Of course, the condition that the series must rapidly converge isn't at all sufficient. But the argument seems to be working well with this condition. $\endgroup$ – Redundant Aunt Jun 13 '15 at 21:48
  • $\begingroup$ Series of reciprocals of integers that rapidly converge are irrational, even though the same is not true of series of arbitrary natural numbers. So the rapidity of convergence is at least a factor. $\endgroup$ – Matt Samuel Jun 13 '15 at 22:40
  • $\begingroup$ Sure, for reciprocals of integers. Even so, one uses a different proof for the second than for the first. $\endgroup$ – André Nicolas Jun 13 '15 at 23:03
  • $\begingroup$ And a variation of this is how Liouville showed the existence of transcendental numbers. $\endgroup$ – marty cohen Jun 14 '15 at 0:24

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