Does $\prod_{k=1}^\infty 1- \frac{1}{k^\alpha}$ converge for $\alpha >1$?

Question

I know that, given $\alpha \in \mathbb{R}$, the infinite product

$$\prod_{k=1}^\infty 1+ \frac{1}{k^\alpha}$$

converges if and only if $\alpha > 1$. My question is whether something similar applies to

$$\prod_{k=2}^\infty 1- \frac{1}{k^\alpha}.$$

Unfortunately, the only useful estimates with logarithms that I could find assume that we are multiplying numbers greater than $1$.

Answer 1

Using the bounds

$$\frac{1}{1-k^\alpha}\le \log\left(1-\frac1{k^\alpha}\right)\le -\frac1{k^\alpha}$$

it is evident that

$$\log\left(\prod_{k=2}^K \left(1-\frac1{k^\alpha}\right)\right)=\sum_{k=2}^K \log\left(1-\frac1{k^\alpha}\right)$$

converges for $\alpha>1$ and diverges otherwise. Can you conclude now?

Answer 2

Hint. Observe that $$ \log \left[\prod_{k=2}^N \left(1- \frac{1}{k^\alpha} \right) \right]=\sum_{k=2}^N\log \left(1- \frac{1}{k^\alpha} \right),\qquad n\ge2, $$ and, as $ k \to \infty$, $$ \log \left(1- \frac{1}{k^\alpha} \right)\sim - \frac{1}{k^\alpha}. $$ Can you take it from here?