Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find the limit (where a is a constant)


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

share|cite|improve this question
up vote 11 down vote accepted

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 $$\exp(x) = 1 + x + \mathcal{O}(x^2)$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.