25
$\begingroup$

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?

$\endgroup$
5
  • 7
    $\begingroup$ +1 for "Harmonacci" $\endgroup$
    – Dirk
    Jan 18, 2013 at 8:08
  • 6
    $\begingroup$ So, in terms of continued fractions, we have $a_{n+1}=[a_{n-1}; a_{n-2},\dots, a_2, a_1, a_0, a_1]$. $\endgroup$ Jan 18, 2013 at 8:59
  • 1
    $\begingroup$ For $a_0=2n+1$, the numerator of the limit $l_{2n+1}$ seems to be $(1\times 3\times...(2n+1))^2\pi$ and its denominator $2^{2n}\times D(l_{2n-1})$, where $D(l_{2n-1})$ is the denominator of $l_{2n-1}$. And it seems that $l_{2n}=2^{2n}/l_{2n-1}$. Maybe a proof by reccurence could be achieved. $\endgroup$ Jan 18, 2013 at 10:05
  • 2
    $\begingroup$ A cute reformulation: start with a unit square, put another next to it on the left, then a 2 x 1/2 along the top and continue always adding a unit area rectangle alternating top and left. What is the limiting ratio of length to height? $\endgroup$ Jan 18, 2013 at 13:11
  • $\begingroup$ Aaron - Cute indeed. $\endgroup$
    – Victor P
    Jan 20, 2013 at 6:24

1 Answer 1

23
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.