Convergence of the infinite product of $\cos(1/\sqrt{i})$ I am trying to prove that the following product is convergent and show what it converges to:

$$\prod_{i=1}^\infty \cos{ \left( \frac{1}{\sqrt{i}} \right)}$$

I have heard that products are convergent if and only if the series of the $\ln$ of the product is convergent:

$$\ln \left[ \prod_{i=1}^\infty \cos{ \left( \frac{1}{\sqrt{i}} \right)} \right] = \sum_{i=1}^\infty \ln \left[ \cos{ \left( \frac{1}{\sqrt{i}} \right)} \right]$$

I have tried to prove that this is convergent in a number of ways to no avail. I believe it must therefore be divergent, but I have been having equally as much trouble proving this. It seems I have to prove that it is divergent using the comparison test; however, I was under the impression that this test only holds for series of positive reals.
In a neighbourhood of the origin,
$$ \log\cos x = -\frac{x^2}{2}-\frac{x^4}{12}-O(x^6) \tag{1}$$
hence
$$ \sum_{k=1}^{N}\log\cos\frac{1}{\sqrt{k}} = -\frac{H_N}{2}+O(1) = -\log\sqrt{N}+O(1)\tag{2} $$
and by exponentiating back we get that
$$ \prod_{k=1}^{N}\cos\left(\frac{1}{\sqrt{k}}\right) \approx \frac{C}{\sqrt{N}}\tag{3} $$
so the value of the given infinite product is just zero.
$\sum_{i=1}^\infty \ln(\cos{(\frac{1}{\sqrt{i}})})$

$\ln(\cos{(\frac{1}{\sqrt{i}})}) = \frac 12\ln((1-\sin^2{(\frac{1}{\sqrt{i}})})$

$\ln((1-\sin^2{(\frac{1}{\sqrt{i}})})&lt;-\frac 1i$

$\sum_{i=1}^\infty \ln(\cos{(\frac{1}{\sqrt{i}})}) &lt; -\frac 12\sum_{i=1}^\infty \frac{1}{i}$

$\sum_{i=1}^\infty \frac{1}{i}$ diverges so $\sum_{i=1}^\infty \ln(\cos{(\frac{1}{\sqrt{i}})})$ diverges.  

$\lim_\limits{n\to\infty}  \sum_\limits{i=1}^n \ln(\cos(\frac{1}{\sqrt i})) = -\infty\\\lim_\limits{n\to\infty}  \exp(\sum_\limits{i=1}^n \ln(\cos{(\frac{1}{\sqrt{i}})})) = 0$