Let $\alpha_n$ be sequence that converges to $\alpha$. I'd like to find such numbers $c, d$ that $\alpha_n - \alpha \approx cn^-d$.
I've found a solution to a similar problem - finding the order of convergence. I'd like to modify it to find the rate instead. Let's take a formula to approximate the order: $ d \approx \frac{ \ln | (\alpha_{n+1} - \alpha_n) / (\alpha_n - \alpha_{n-1})| } { \ln | (\alpha_n - \alpha_{n-1}) / (\alpha_{n - 1} - \alpha_{n-2}) | } $.
It is derived from $|\alpha_n - \alpha| \approx c|\alpha_{n-1} - \alpha|^d$
Since I have $|\alpha_n - \alpha| \approx cn^d$, the formula for the rate would be (if I'm not mistaken): $d \approx \frac{\ln |(\alpha_{n + 1} - \alpha_n) / (\alpha_n - \alpha_{n - 1})|}{\ln ((n + 1) / n)}$.
I tried testing it on some examples, but it seems like either I had made a mistake in the formula or it's not very accurate. Suprisingly (at least for me), $d$ calculated using the first formula for the order was almost perfect. Am I doing something wrong?
Edit: I've checked that after putting the value of $\alpha$ in the formula ($d \approx \frac{ln|(\alpha_{n+1} - \alpha)/(\alpha_n - \alpha)}{ln((n+1)/n)}$) it works pretty well. I assume then the problem lies here. Is there any other way to get rid of $\alpha$ value?
Let $b_n = cn^d$, then $$\frac {b_{n+1}}{b_n} = \left(1 + \frac 1n\right)^d = 1 + d\frac 1n + \frac {d(d-1)}2\left(\frac 1n\right)^2 + ...$$
and so $$d = \lim_n n\left(\frac {b_{n+1}}{b_n} - 1\right)$$
Translating to $\alpha_n$, this becomes $$d = \lim_n n\left(\frac {\alpha_{n+1}-\alpha}{\alpha_n - \alpha} - 1\right)=\lim_n n\left(\frac {\alpha_{n+1}-\alpha_n}{\alpha_n - \alpha}\right)$$
Of course, once you have $d$, $c$ is just $$c = \lim_n\frac{\alpha_n - \alpha}{n^d}$$