User felix janda - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T17:39:56Z http://mathoverflow.net/feeds/user/30755 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119241/harmonacci-recurrence-and-identities-for-pi/119251#119251 Answer by Felix Janda for "Harmonacci" recurrence and identities for $\pi$ Felix Janda 2013-01-18T10:42:48Z 2013-01-18T11:57:45Z <p>The sequence $a_n$ is closely related to the Wallis product <code>$$a'_n = \prod_{i = 1}^n \left(\frac{2i}{2i - 1} \frac{2i}{2i + 1}\right),$$</code> which converges to $\pi/2$ as $n$ goes to infinity. Namely, we have <code>$$a'_n = a_{n + 1} \cdot \frac{2n}{2n + 1}$$</code>. 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 $b_n$.</p> <p>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.</p>