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I read here that $\sum_{n=1}^\infty \frac{1}{d_n}$ is irrational if $d_{n+1} > d_{n}^2$ for all $n > N_0$.

Can we prove $\pi$, $e$ or some other numbers irrational by creating a series for them that converges like this?

I looked on wikipedia if there were some already and found Ramanujans series for $\pi$: $$\frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}$$ and (if I did the inequality correct) even this is not fast enough for irrationality of $\pi$!

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If applicable at all, the Ramanujan series would show that $\pi\sqrt 2$ is irrational, thus still leaving the possibility that $\pi$ itself is rational. – Hagen von Eitzen Jan 20 '13 at 13:10
Just be careful, because the $d_i$ in the above theorem need to be integers – Thomas Andrews Jan 20 '13 at 13:10
@HagenvonEitzen, good point! Thank you for mentioning that. – user58512 Jan 20 '13 at 13:10
It's not hard to prove the irrationality of $e$ by looking at the rate of convergence of $\sum 1/n!$. This is done in Rudin's Principles, for example. The quadratic estimate you quoted is asking more than is necessary. – user53153 Jan 20 '13 at 18:57
@5PM, yeah that proof is very nice. I am interested in whether this particular irrationality criterion has value, and as you mention it is not strong enough to apply to the sequence n!. I think the answer to my question is that it doesn't really have a use, non-contrived numbers don't seem to have such fast converging series. – user58512 Jan 20 '13 at 19:04

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