How would you solve the following equation for $p>0$?
$$\text{argmin}_{\alpha} \max_{i\in\{1,2,3,\ldots\}} \left(1-\alpha \frac{1}{i^p}\right)^2 \frac{1}{i}$$
This comes down to minimizing an envelope of quadratics. For $p=1$, the answer is determined by intersection of first and fourth curves.
$$\text{argmin}_{\alpha} \max_{i\in\{1,2,3,\ldots\}} \left(1-\alpha \frac{1}{i}\right)^2 \frac{1}{i}=\frac{4}{3}$$
For $p=2$
$$\text{argmin}_{\alpha} \max_{i\in\{1,2,3,\ldots} \left(1-\alpha \frac{1}{i^2}\right)^2 \frac{1}{i}=\frac{9}{121} \left(4 \sqrt{3}+13\right)$$
obtainable from intersection of first and third curves.
What about more general $p$?