Fast convergence implies higher likelihood of irrationality/transcendentalism? While browsing old questions I came across this one about evaluating $\displaystyle\sum_{n=0}^{\infty} \frac{n!}{(n^2)!}$, and two of the comments in particular caught my eye.  They're reproduced here just in case:

Because the convergence is so rapid, the limit is apt to be a transcendental number. – hardmath

and

@hardmath: The convergence is certainly fast enough to show that the number is irrational. However, one would need each term to be smaller than the square of the previous term (roughly) to imply transcendence, and that isn't the case here. – Greg Martin

I didn't know that fast convergence is related to irrationality and transcendentalism.  I looked it up on the web and didn't find much that explains this relationship.  Can anyone shed light on it or provide references that discuss it?
 A: A starting point might be 

Lagrange's theorem: If $\left|\alpha-\frac{p}{q}\right|\leq\frac{1}{2q^2}$, $\frac{p}{q}$ is
  a convergent of the continued fraction of $\alpha$.

Since a rational number only has a finite number of convergents, the existence of a sequence of rational numbers $\left\{\frac{p_n}{q_n}\right\}_{n\geq 1}$ with the property
$$ \left|\alpha-\frac{p_n}{q_n}\right|\leq \frac{1}{q_n^c} $$
for some $c>2$ implies the irrationality of $\alpha$. In particular, many irrationality proofs use Beuker integrals for constructing sequences with such a property. For $\zeta(2)$ and $\zeta(3)$, such sequences are given by linear combinations (with coefficients in $\mathbb{Q}$) of the partial sums of the fast-convergent series
$$ \sum_{n\geq 1}\frac{3}{n^2\binom{2n}{n}},\qquad \sum_{n\geq 1}\frac{5(-1)^{n+1}}{2n^2\binom{2n}{n}}.$$
Roughly speaking, a fast-convergent series with rational terms (with some additional arithmetic constraint on its partial sums) is likely to converge to an irrational number.
