Let ${\displaystyle a_{0}=0}$and ${\displaystyle a_{n+1}={\sqrt {A+Ba_{n}}}.}$Then, assuming convergence, we have ${\displaystyle a_{\infty }=\ell ={\frac {B+{\sqrt {B^{2}+4A}}}{2}}.}$Thus, for ${\displaystyle A=B=1}$we have ${\displaystyle \ell =\phi ,}$for instance. Now, for ${\displaystyle A=B={\frac {1}{2}}}$we have ${\displaystyle \ell =1,}$and ${\displaystyle \lim _{n\to \infty }{\color {red}2}^{n}{\sqrt {\frac {\ell -a_{n}}{2}}}={\frac {\pi }{4}}.}$My question would be with what constant to replace ${\displaystyle {\color {red}2}}$in general, for different values of A and B, so that the limit in question should converge to a finite non-zero quantity. In other words, if ${\displaystyle f\left({\frac {1}{2}},{\frac {1}{2}}\right)=2,}$what is the general formula for ${\displaystyle f(A,B)}$ ? Thank you. — 79.118.171.25 (talk) 22:57, 22 June 2015 (UTC)[reply]

Apparently, ${\displaystyle f(1,1)={\sqrt {1+{\sqrt {5}}}}~,}$and the limit in question is the square root of the Paris constant. — 79.118.171.25 (talk) 03:07, 23 June 2015 (UTC)[reply]

I get ${\displaystyle f(A,B)={\sqrt {2\ell /B}}}$. My idea is to let ${\displaystyle x_{n}=\ell -a_{n}}$and write its recurrence formula. The behavior for small ${\displaystyle x}$is dictated by the linear term of its Maclaurin series. ${\displaystyle \ell -{\sqrt {A+B(\ell -x)}}=0+{\frac {B}{2\ell }}x+...}$From there it's not hard to understand the behavior of ${\displaystyle {\sqrt {\frac {x_{n}}{2}}}}$and find ${\displaystyle f(A,B)}$. Egnau (talk) 03:40, 23 June 2015 (UTC)[reply]

I arrived just these past few minutes at the same conclusion, and wanted to post it, but was unable to connect. :-) Thanks ! — 79.113.226.120 (talk) 03:53, 23 June 2015 (UTC)[reply]

And in general, for ${\displaystyle a_{n+1}={\sqrt[{m}]{A+Ba_{n}}}}$we have ${\displaystyle f_{m}(A,B)={\sqrt {\frac {m~\lambda ^{m-1}}{B}}}~,}$where ${\displaystyle \lambda }$is a root of ${\displaystyle t^{m}=A+Bt.}$— 79.113.226.120 (talk) 10:08, 23 June 2015 (UTC)[reply]

Resolved