Quadratic convergence of $b_k=\frac{1}{2^{2^k}}$

Question

Why does the series $b_k=\frac{1}{2^{2^k}}$ converges quadratically? My calculations are

$\lim_{k\rightarrow \infty}|\frac{\frac{1}{2^{2^{k+1}}}}{^\frac{1}{2^{2^k}}}|=|\frac{1}{2^{2^k}}|\rightarrow 0$ Because of $\lim_{k\rightarrow \infty} 2^{2^k}\rightarrow \infty$

I searched on Wikipedia and it says that it converges quadratically and not linearly.

I don't understand where, however, I am wrong in my reasoning.

Can someone tell me?

Answer 1

Because $b_{k+1}=(b_k)^2$.

As you found, the simple quotient converges to zero, so that the convergence is at least super-linear.

In continuation, $c_k=2^{-3^k}$ would converge cubically,...