There evidently exist infinite series that converge to their limits with arbitrary speed. For example, for each $\alpha>0$, the series
$$ \sum_{j=1}^n j^{-\alpha-1} \to \sum_{j=1}^\infty j^{-\alpha-1} $$ with rate of convergence can be computed by comparison with an integral, and is given by
$$ \sum_{j=1}^\infty j^{-\alpha-1} - \sum_{j=1}^n j^{-\alpha-1} =\sum_{j=n+1}^\infty j^{-\alpha-1} =O(n^{-\alpha}). $$ In other words, there exist series that converge to their limits with speed $n^{-\alpha}$ for any $\alpha>0$.
My question is whether the same hold for infinite products. I.e., for any $\alpha>0$, does there exist a sequence $a_j$ satisfying $0 < a_j <1$, and $0 < \prod_{j=1}^\infty a_j < 1$, so that $$ \prod_{j=1}^n a_j -\prod_{j=1}^\infty a_j = O(n^{-\alpha})? $$
I assume the answer is yes, but I cannot see how. Any help much appreciated!
As $$\prod_{j=1}^n a_j-\prod_{j=1}^\infty a_j=\prod_{j=1}^\infty a_j\,\left (\prod_{j=n+1}^\infty a_j^{-1}-1\right )$$ The question is equivalent to $$1+cn^{-a}\le \prod_{j=n+1}^\infty a_j^{-1}\le 1+Cn^{-a }$$ Thus $$\log(1+cn^{-a})\le \sum_{j=n+1}^\infty \log(a_j^{-1})\le \log(1+Cn^{-a})$$ As $${1\over 2}x\le\log (1+x)\le x, \qquad 0<x<1$$ the last inequality is implied by $$cn^{-a}\le \sum_{j=n+1}^\infty \log(a_j^{-1})\le {1\over 2}Cn^{-a}$$ The last condition is satisfied when $\log(a_j^{-1})=j^{-a-1}$, i e. $$a_j=\exp\left (-j^{-a-1} \right ) $$