Finding an approximation for the difference of $a_n = \frac{1}{1+a_{n-1}}$ and it's limit. I've got the recurrence
$\displaystyle{a_{n} = {1 \over 1 + a_{n - {\tiny 1}}},\ }$
for $0 < a_{0} < 1 $ which has the solution
$\displaystyle{\alpha = {\,\sqrt{\, 5\,}\, - 1 \over 2}}$
I am now trying to find an approximate formula for $b_n = a_n - \alpha$.
I can see that $a_n$ is monotonically decreasing, and therefore any $a_n$ could be written as a geometric series...but I don't see how to use that fact.
 A: For every $n$, $$b_{n+1}=-\frac{\alpha b_n}{1+a_n},$$ hence the sequence $(b_n)$ has alternating signs in the sense that $b_nb_{n+1}\leqslant0$. Furthermore, $1+a_n\geqslant1$ hence $|b_{n+1}|\leqslant\alpha\cdot|b_{n}|$ for every $n$ hence $$|b_{n}|\leqslant\alpha^n\cdot|b_0|\to0.
$$
However, the true order of convergence when $n\to\infty$ is not $\alpha^n$ but $\alpha^{2n}$. To see this, note that $1+a_n\to1+\alpha=1/\alpha$ hence $b_{n+1}/b_n\to-\alpha^2$, and, when $n\to\infty$, $$|b_{n}|=\alpha^{2n+o(n)}.$$ With more work, one can actually refine this to $|b_{n}|=\alpha^{2n+O(1)}$, and even to the fact that there exists some positive $c(a_0)$ such that, when $n\to\infty$, $$(-1)^n\alpha^{-2n}\cdot b_{n}\to c(a_0).$$ And, to be quite specific, $$c(a_0)=\sqrt5\cdot\frac{a_0-\alpha}{a_0+\alpha+1}.$$
A: Start from $a_n = \alpha - b_n$:
$$
a_{n+1} = \frac{1}{1+\alpha - b_n} = \frac{1}{\rho - b_n}
$$
with $\rho = \frac{\sqrt{5}+1}{2} = 1 + \alpha = \frac{1}{\alpha}$.
$$
b_{n+1} =  \alpha - \frac{1}{\rho - b_n}
=\frac{\alpha \rho - \alpha b_n -1}{\rho-b_n} = \left( -\frac{\alpha}{\rho} \right)b_n 
\left[1 - \frac{b_n^2}{\rho - b_n} \right]
$$
Thus far, this is exact, but now we will simplify by saying that 
$$
\frac{b_n^2}{\rho - b_n} \ll 1
$$
From this we see that the $b_n$ osscilate positive to negative, that 
$$ b_n \approx b_0 (-1)^n \alpha^{2n} = (\alpha-a_0) (-1)^n \alpha^{2n} $$ 
where the approximation is quite good for small $b_0$, and that for any starting $a_0$ between 0 and 1 
$$ |b_n| < \alpha^{2n}$$
