"Harmonacci" recurrence and identities for $\pi$ - MathOverflow

"Harmonacci" recurrence and identities for $\pi$

2013-01-18T07:11:49Z

While playing with something totally irrelevant I stumbled upon the recurrence: $$a_{n+1} = \frac{1}{a_n} + a_{n-1}$$

It turns out that given $a_0 = 1, a_1 = 1$,

$$lim \frac{a_{2n}}{a_{2n-1}} = \frac{\pi}{2}$$

I have a very crude idea (or rather a hint) on proving it (the iterations sort of unfold into a sort of Viete product, which is sort of expected), but my technique is rusty at best.

With different initial conditions, things start getting really scary, for example $ a_0 = 2, 3, 4, 5 $ yield $\frac{8}{\pi}, \frac{9\pi}{8}, \frac{128}{9\pi}, \frac{225\pi}{128}$ respectively.

So, the questions are: Is it a known fact? If so, where can I read more on it? If not, may anybody help me to prove/disprove it? Does it mean anything?

Answer by Felix Janda for "Harmonacci" recurrence and identities for $\pi$

2013-01-18T10:42:48Z

The sequence $a_n$ is closely related to the Wallis product $$a'_n = \prod_{i = 1}^n \left(\frac{2i}{2i - 1} \frac{2i}{2i + 1}\right),$$ which converges to $\pi/2$ as $n$ goes to infinity. Namely, we have $$a'_n = a_{n + 1} \cdot \frac{2n}{2n + 1}$$. This could be proven by induction or maybe more easily by defining $b_n = a_n a_{n - 1}$ and noticing that the recursion for $a_n$ implies the (very simple) recursion $$b_{n + 1} = 1 + b_n$$ for $b_n$ and expressing $a_n$ in terms of the $b_n$.

For more general values of $a_0$ one gets similar formulas for $a_n$ as (up to a factor converging to 1) a Wallis product or inverse of a Wallis product where a few of the lower terms in the product are missing.