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A less known product formula for $\pi$, due to Sondow, is the following:

$$ \frac{\pi}{2}= \left(\frac{2}{1}\right)^{1/2} \left(\frac{2^2}{1\cdot3}\right)^{1/4} \left(\frac{2^3\cdot4}{1\cdot3^3}\right)^{1/8} \left(\frac{2^4\cdot4^4}{1\cdot3^6\cdot5}\right)^{1/16} \ldots $$

Is it possible to prove the irrationality of $\pi$ based on this formula, using some sort of convergence acceleration technique leading to a good irrationality measure?

Blue
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Klangen
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  • What makes you think it could be possible?, i mean, what's special about that formula? – Aldama Aug 11 '21 at 13:02
  • @Aldama It converges, therefore it has a convergence rate. The convergence rate can be computed. If it is sufficiently quick, we can use Dirichlet's Theorem to prove irrationality. – Klangen Aug 11 '21 at 13:36
  • @Klangen Which Dirichlet theorem? – FShrike Aug 12 '21 at 09:03
  • @FShrike This one: https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem – Klangen Aug 12 '21 at 09:21
  • every formula that gives $\pi$ can be used to prove everything about $\pi$. However maybe the way is not straighforward –  Aug 14 '21 at 07:15

1 Answers1

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Dirichlet’s approximation theorem can’t be applied here, as this product formula doesn’t give a sequence of rational approximations due to the fractional exponents.

We can show that this product formula is equivalent to the Wallis product, which does give you a sequence of rational approximations:

Let $a_n(x)$ be the product of the first $n$ terms of

$$\left(\frac{x + 1}{x}\right)^{1/2} \left(\frac{(x + 1)^2}{x(x + 2)}\right)^{1/4} \left(\frac{(x + 1)^3(x + 3)}{x(x + 2)^3}\right)^{1/8} \left(\frac{(x + 1)^4(x + 3)^4}{x(x + 2)^6(x + 4)}\right)^{1/16} \cdots.$$

Then we can see that $a_n(x)$ satisfies this recurrence:

$$a_0(x) = 1, \quad a_{n+1}(x) = \left(\frac{x + 1}{x}\right)^{1/2} \left(\frac{a_n(x)}{a_n(x + 1)}\right)^{1/2}.$$

In the limit as $n \to \infty$, this simplifies to

$$\quad a(x) = \frac{(x + 1)}{x a(x + 1)},$$

and since $\lim_{x \to \infty} a(x) = 1$, we can write

$$a(1) = \frac21 \cdot \frac1{a(2)} = \frac21 \cdot \frac23 \cdot a(3) = \cdots = \frac21 \cdot \frac23 \cdot \frac43 \cdot \frac45 \cdot \frac65 \cdot \frac67 \cdots.$$

However, since the Wallis product doesn’t qualify as lesser-known, I’d imagine that an irrationality measure proof based on it would be well known if it existed.

Anders Kaseorg
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    There are results out there for infinite products of the form $\Pi_{n=1}^\infty (1+1/a_n)$, to the effect that this product is irrational if $a_n$ is a positive integer sequence which increases fast enough (c.f. the remarks on the first page of the article here). But the Wallis product corresponds to $a_n=4n^2-1$, and this doesn't anywhere near fast enough to apply such results. (It badly fails to satisfy the premises in the linked article, for instance.) – Semiclassical Aug 14 '21 at 09:26