<s>[INST] Sufficient condition for infinite product being $>0$? I'm trying to solve the following problem:

When is $\prod_{k=1}^{\infty}\frac{\left(\frac{1}{2}-\frac{1}{2}k^{-\alpha}\right)^{1/2}+\left(\frac{1}{2}+\frac{1}{2}k^{-\alpha}\right)^{1/2}}{2^{1/2}}>0$, for $\alpha >0$?


Since $\frac{1}{2}-\frac{1}{2}k^{-\alpha}$ converges to $\frac{1}{2}$ as $k\to\infty$, the fraction $\frac{\left(\frac{1}{2}-\frac{1}{2}k^{-\alpha}\right)^{1/2}+\left(\frac{1}{2}+\frac{1}{2}k^{-\alpha}\right)^{1/2}}{2^{1/2}}$ converges to $1$ but I'm having serious trouble with the infinite product. How exactly do I use this observation to find sufficient conditions for $\alpha>0$?
 [/INST] Taking the logarithm, you see that the product is $>0$ if and only if
$$ \sum_{k=1}^{+\infty}\log\left(\frac{\sqrt{1-k^{-\alpha}}+\sqrt{1+k^{-\alpha}}}{2}\right) $$
converges.
As $k\rightarrow +\infty$,
$$ \begin{aligned} \log\left(\frac{\sqrt{1-k^{-\alpha}}+\sqrt{1+k^{-\alpha}}}{2}\right) &= \log\left(1+\frac{1}{2k^{2\alpha}}+o\left(\frac{1}{k^{2\alpha}}\right)\right)\sim\frac{1}{2k^{2\alpha}}
\end{aligned} $$
Therefore the sum converges if and only if $\alpha>\frac{1}{2}$, which means that the product is $>0$ if and only if $\alpha>\frac{1}{2}$.
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