In the course of exploring the values of a family of games, I needed to solve a non-linear recursion. The approach taken was an asymptotic expansion in $n$, the index in this family of the game being valued.
The equation in question is $$ a_n = \frac{3a_{n-1}-a_{n-2}+4a_{n-1}a_{n-2}}{2+6a_{n-1}-2a_{n-2}+8a_{n-1}a_{n-2}} $$ with $a_1 = a_2 = \frac16$.
Having done some perturbation theory I have been able to obtain a starting point $$ a_n = \frac{1}{4\sqrt{n}} +\frac{5}{8n} + \frac{f_n}{n} $$ where as long as $f_n$ goes to zero satisfies the recursion equation up to and including order $n^{-2}$. $f_n$ cannot be identically zero, since then the recursion will read $\frac{9}{64}n^{-5/2} = 0$.
That $\frac{9}{64}n^{-5/2}$ has proven very hard to cancel by trying various forms for $a_n$. For example, if you replace the $\frac{1}{4\sqrt{n}}$ by a sum of more complicated terms containing adjustable parameters, and adjust the parameters to force cancellation at orders up to $n^{-2}$, the contribution to the $n^{-5/2}$ term seem always to cancel out, leaving that $\frac{9}{64}n^{-5/2} $ intact. I've tried a large number of times along these lines.
A bit of work with the series (expanded in $\frac{1}{\sqrt{n}}$) representing the recursion, expressed in terms of $f_n$, leads me to try an assumption that $$ f_{n-1} = f_n(1+k/n+r/n^2\ldots)\\ f_{n-2} = f_n(1+2k/n+s/n^2\ldots)\\ $$ which allows the $f_n/n^2$ term to cancel the $\frac{9}{64}n^{-5/2} $ if $$ f_n = \frac{51}{160\sqrt{n}} + o(n^{-1/2} ) $$ and $$ k = \frac{47}{34} $$
But that involves finding some $f(n)$ that falls as $\frac{a}{\sqrt{n}}$ while the expansion of $f(n-1)$ is $f(n)(1+ \frac{47}{34n} + \ldots)$. Typical functions, such as $n^{-1/2}$, yield $\frac12$ in place of that $\frac{47}{34}$.
So the question:
Can there be any $f(n)$ such that $f(n) \to \frac{a}{\sqrt{n}}$ as $n->\infty$, but $\frac{f(n-1)}{f(n)}\to 1+\frac{k}{n} + O(1/n^2)$ with $k\neq\frac12$?
And if there cannot be, can you prove that?
I'm providing an answer so as to get this off the "unanswered" list.
Because there is no way to cancel the $-\frac9{64}n^{-5/2}$ using an ordinary exmansion in powers of $n^{-1/2}$ I ended up having to move outside that box. After a staggering number of tries using expressions like $a_n = \frac{\alpha}{\sqrt{n+\beta}} + \frac{\gamma}{n+\delta} $, which all led to the same failure to cancel the $-\frac9{64}n^{-5/2}$, I finally looked at multiplying the next term by the harmonic number $H_n$.
In this way, it turns out that with the next term, the perturbation series is $$ a_n = \frac1{4\sqrt{n}}+\frac5{8n}-\frac9{32}\frac{H_n}{n^{3/2}} $$ where $$H_n = \sum_{k=1}^n \frac1k$$
Since this brough my agreement with numerical calculations into the part per trillion range wor modest values of $n$, this is good enough to satisfy my obsession with going to the next term.