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On the OEIS Wiki immediately after the formula $$\sqrt{\frac{\pi e}{2}}=\frac{1}{1+\mathrm{K}_{i=1}^{\infty}{\frac{i}{1}}}+\sum_{n=0}^{\infty}{\frac{1}{(2n+1)!!}}$$ (where I am using $\mathrm{K}$ as in the third notation here for continued fractions) it is written "Since it is not known whether $\pi e$ is irrational or not, it is thus not known whether $\sqrt{\pi e/2}$  is transcendental or not (although it is obviously irrational)."

How can the irrationality of $\sqrt{\pi e/2}$ be derived from the formula?

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  • $\begingroup$ What on earth does the big $K$ mean?……Oh. $\endgroup$ – user477343 Feb 2 '18 at 14:11
  • $\begingroup$ I don't see why that would be obvious, and it would also imply that $\pi e$ is not of the form $\frac{2p^2}{q^2}$ for integer $p, q$. $\endgroup$ – orlp Feb 2 '18 at 15:26
  • $\begingroup$ I think this sum is due to Ramanujan, and there already is another question similar to this. $\endgroup$ – Crescendo Feb 2 '18 at 15:45
  • $\begingroup$ Perhaps this may be of use: math.stackexchange.com/questions/833920/… $\endgroup$ – Crescendo Feb 2 '18 at 15:46
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I see nothing obvious (from this particular expression, discounting the other things known about $\pi$ and $e$).

The continued fraction is general, so the fact that it doesn't terminate means nothing (unlike for simple continued fractions).

However, the series represents the Engel expansion of some number $x$. The Engel expansion of a real number is unique. It is also known that the expansion is finite (terminating) iff the number is rational.

Thus:

$$x=\sum_{n=0}^{\infty}{\frac{1}{(2n+1)!!}}=1+\frac{1}{3}+\frac{1}{3 \cdot 5}+\frac{1}{3 \cdot 5 \cdot 7}+\cdots$$

is irrational.

(In the same way, it is 'obvious' that $e$ is irrational, going by its Engel expansion, which is the same as one of its usual definitions as $\sum_{n=0}^\infty \frac{1}{n!}$).

Of course, the fact that one part of the sum is irrational, doesn't mean that the whole sum is, since, in principle, the continued fraction could have the form $\frac{p}{q}-x$, where $p,q \in \mathbb{N}$.

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