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.

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    $\begingroup$ The partial products are positive and decreasing aren't they? $\endgroup$ – zhw. Dec 7 '16 at 22:18
  • $\begingroup$ @zhw. Yes, I believe so. We haven't learned much about the convergence of products in my class, so I am having trouble understanding this.. Is this enough to prove convergence? $\endgroup$ – user3495690 Dec 7 '16 at 22:20
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    $\begingroup$ The product diverges to $0$. When one looks at infinite products, $0$ is an annoying number. It's analogous to $+\infty$ and $-\infty$ for series. It has its uses, but one says that a product diverges if the sequence of partial products tends to $0$, but no factor is $0$. $\endgroup$ – Daniel Fischer Dec 7 '16 at 22:23
  • $\begingroup$ @DanielFischer Thank you! I had read that, but wasn't sure if it was a 100% reliable method. +1 :) $\endgroup$ – user3495690 Dec 7 '16 at 22:27

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.


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

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

$\ln((1-\sin^2{(\frac{1}{\sqrt{i}})})<-\frac 1i$

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

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

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

  • $\begingroup$ $\lim_\limits{n\to\infty} \sum_{i=1}^n \ln(\cos(\frac{1}{\sqrt i})) = -\infty\\\lim_\limits{n\to\infty} \exp(\sum_{i=1}^n \ln(\cos{(\frac{1}{\sqrt{i}})})) = 0$ $\endgroup$ – Doug M Dec 7 '16 at 22:37
  • $\begingroup$ But doesn't the comparison test for divergent series only hold for series of nonnegative terms? $\endgroup$ – user3495690 Dec 8 '16 at 17:41
  • $\begingroup$ Also, I didn't look through your work very closely until now, but it seems you are also using the comparison test incorrectly. For $-1/i$ to imply divergence, it must be less than $\ln(\cos(1/\sqrt(i)))$ $\endgroup$ – user3495690 Dec 9 '16 at 1:41
  • $\begingroup$ $|\ln\cos \frac 1{\sqrt i} | > | \frac 1i|$ and all the terms have the same sign. Which proves divergence. Or, $\ln\cos \frac 1{\sqrt i} < - \frac 1i$ which is the same thing. $\endgroup$ – Doug M Dec 9 '16 at 17:18
  • $\begingroup$ I see what you're saying now, but wouldn't this only prove it's not absolutely convergent? (Meaning it could still be conditionally convergent.) I'm aware that it is in fact divergent, but even so.. $\endgroup$ – user3495690 Dec 10 '16 at 5:59

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