$\sqrt{\frac{\pi e}{2}}=\frac{1}{1+\mathrm{K}_{i=1}^{\infty}{\frac{i}{1}}}+\sum_{n=0}^{\infty}{\frac{1}{(2n+1)!!}}$ implies $\sqrt{\pi e/2}\notin Q$?

185 Views Asked by At

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?

1

There are 1 best solutions below

0
On

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 $p,q \in \mathbb{N}$.