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.

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, $$c(a_0)=\sqrt5\cdot\frac{a_0-\alpha}{a_0+\alpha+1}.$$
