Need help with gradient descent convergence calculation We are given $\alpha=1-8\eta$, where $\eta$ is the step size. We are also given the following ($z$ is irrelevant, but basically it's the output of the algorithm at step $k$):

We are asked what values of $\eta$ in the set $\{0.1,0.11,0.12,0.13,0.14,0.15\}$ make the gradient descent take the most steps before convergence. The answer is $\eta=0.1$ and $\eta=0.15$.
In a previous question, we find out that $\eta=0.125$ causes the gradient to converge in a single step/without oscillating, since according to the information convergence without oscillation happen when $1>\alpha\ge0$, therefore when $\alpha=1=1-8\eta$, $\eta=0.125$.
Therefore, the solution goes, since these two $\eta$ are the furthest from $0.125$, they would make the descent the slowest/takes the most steps.
I do not understand this; if we plug the values from that set into $\alpha=1-8\eta$, we get $$
\begin{array}{c|lcr}
\alpha & \eta  \\
\hline
0.2 & 0.1  \\
0.12 & 0.11  \\
0.04 & 0.12\\
0 & 0.125\\
-0.04 & 0.13\\
-0.12 & 0.14\\
-0.2 & 0.15 
\end{array}
$$
According to the table, convergence occurs without oscillation when $1>\alpha\ge 0$, so shouldn't it converge immediately when $\eta$ is within $[0.1,0.125]$?

Excerpts from the relevant MIT notes have been attached below for reference:


 A: They have rewritten the problem by changing variables to $z^{(k)}$, so that the global minimum of the function occurs at $z^{(k)}=0$, and each update of gradient descent is just multiplying $z^{(k)}$ by a constant. If that constant is zero, then it converges in one step. If the constant is positive but less than one, it converges without oscillating, but it never "reaches" the global minimum. Your gradient descent algorithm would need some stopping criteria. A typical choice is when the norm of the gradient is below some positive threshold.
In this problem, the gradient is proportional to $z^{(k)}$, in particular $\nabla_x f(x^{(k)}) = 8z^{(k)}$, so a threshold on the gradient is the same as a (different) threshold on the size of $z^{(k)}$.
The smaller the positive constant, the fewer steps the algorithm would take before reaching the threshold.
Of course, "number of steps" is an integer, so for each positive integer, there would be a range of values that all took that number of steps to reach the threshold.
