# Find $\lim_{n\to\infty}\prod_{k=1}^n\cos\left(\frac{ka}{n\sqrt{n}}\right)$

Find the limit (where a is a constant)

$\lim_{n\to\infty}\prod_{k=1}^n\cos\left(\frac{ka}{n\sqrt{n}}\right)$

I think the answer is $1-a^2/6$

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Let $$f(n) = \prod_{k=1}^n \cos \left(\dfrac{ka}{n \sqrt{n}}\right)$$

$$g(n) = \log (f(n)) = \sum_{k=1}^{n} \log \left(\cos \left(\dfrac{ka}{n \sqrt{n}}\right) \right) = \sum_{k=1}^{n} \log \left(1 - \dfrac{\left(\dfrac{ka}{n \sqrt{n}}\right)^2}2 + \mathcal{O} \left( \dfrac{k^4}{n^6}\right)\right)$$

$$\log \left(1 - \dfrac{\left(\dfrac{ka}{n \sqrt{n}}\right)^2}2 + \mathcal{O} \left( \dfrac{k^4}{n^6}\right)\right) = -\left(\dfrac{\left(\dfrac{ka}{n \sqrt{n}}\right)^2}2 + \mathcal{O} \left( \dfrac{k^4}{n^6}\right)\right) + \mathcal{O} \left(\dfrac{k^4}{n^6} \right)$$

$$\sum_{k=1}^{n} \log \left(1 - \dfrac{\left(\dfrac{ka}{n \sqrt{n}}\right)^2}2 + \mathcal{O} \left( \dfrac{k^4}{n^6}\right)\right) = \sum_{k=1}^{n} \left( -\dfrac{\left(\dfrac{ka}{n \sqrt{n}}\right)^2}2 + \mathcal{O} \left( \dfrac{k^4}{n^6}\right)\right)\\ = -\dfrac{a^2}{2n^3} \dfrac{n(n+1)(2n+1)}{6} + \mathcal{O}(1/n)$$

$$\lim_{n \to \infty }\sum_{k=1}^{n} \log \left(1 - \dfrac{\left(\dfrac{ka}{n \sqrt{n}}\right)^2}2 + \mathcal{O} \left( \dfrac{k^4}{n^6}\right)\right) = -\dfrac{a^2}{6}$$

Hence, $$\prod_{k=1}^{\infty} \cos \left(\dfrac{ka}{n \sqrt{n}}\right) = \exp(-a^2/6)$$

The solution you have $1-a^2/6$ is a first order approximation to $\exp(-a^2/6)$ since $$\exp(x) = 1 + x + \mathcal{O}(x^2)$$

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My argument is similar (and I hope a little cleaner) (+1) –  robjohn Oct 19 '12 at 2:32
@robjohn Actually, I do not see a difference between your answer and mine. :) –  user17762 Oct 19 '12 at 2:34
I had not read yours before writing up mine. Sorry. –  robjohn Oct 19 '12 at 3:28