Show that $\lim_{n\to \infty}\cos\frac{a}{n\sqrt n}\cos\frac{2a}{n\sqrt n}\cdots \cos\frac{na}{n\sqrt n} = e^{-a^2/6}$ for any constant $a$ For any constant $a$, find the value of
$$\lim_{n\to \infty}\cos{\left(\frac{a}{n\sqrt{n}}\right)}\cos{\left(\frac{2a}{n\sqrt{n}}\right)} \dotsm \cos{\left(\frac{na}{n\sqrt{n}}\right)}$$ 
I figured out that limit of the sequence is $e^{-a^2/6}$    However, I still couldn't solve this question.. please help me 
I didn't know whether it would help or not, but it might relative to $\ln$ ..
 A: Consider the logarithm:
$$ \log\prod_{k=1}^n \cos{\frac{ka}{n^{3/2}}} = \sum_{k=1}^n \log{\cos{\frac{ka}{n^{3/2}}}}. $$
Expanding the summand as a power series for $n \to \infty$ gives
$$ \log{\cos{\frac{ka}{n^{3/2}}}} = \log{\left( 1-\frac{k^2a^2}{2n^3}+o(n^{-1}) \right)} = -\frac{k^2a^2}{2n^3}+o(n^{-1}) $$
since $k\leq n$. Now, summing this up gives
$$ \sum_{k=1}^n \log{\cos{\frac{ka}{n^{3/2}}}} = -\frac{a^2}{2n^3} \sum_{k=1}^n k^2 + o(1), $$
since the sum of $n$ $o(n^{-1})$ terms is $o(1)$. We know that
$$ \sum_{k=1}^n k^2 = \frac{n^3}{3}+o(n^3) $$
(either by knowing the actual formula or by thinking about approximating the volume of a stepped pyramid by a smooth one), and then we end up with
$$ \log\prod_{k=1}^n \cos{\frac{ka}{n^{3/2}}} = -\frac{a^2}{2n^3} \frac{n^3}{3} + n^{-3}o(n^3)+o(1) = -\frac{a^2}{6}+o(1), $$
so the limit is $-a^2/6$. Exponentiating both sides gives the result.

One might be concerned about the validity of these expansions for large $a$. We can throw away finitely many of the factors without affecting the result, and we can start with a value of $n$ as large as we like so we can begin with the assumption that $akn^{-3/2}$ is always small enough for this to work, and the rest goes through with only minor adjustments.
A: $\frac{d}{dz}\log\cos(z)=-\tan(z)$ and $\tan(z)$ is a convex function on the interval $[0,1]$. In particular
$$ \forall z\in[0,1],\quad z\leq\tan(z)\leq z+z^3$$
leads to
$$ \forall z\in[0,1],\qquad \frac{z^2}{2} \leq -\log\cos z \leq \frac{z^2}{2}+\frac{z^4}{4} $$
and for any $a\in\mathbb{R}$ we have that $\left|\frac{a}{\sqrt{n}}\right|\in[0,1]$ for any $n$ large enough. With such assumption we have
$$ \sum_{k=1}^{n}\log\cos\left(\frac{ak}{n\sqrt{n}}\right) \leq -\frac{1}{2}\sum_{k=1}^{n}\frac{k^2 a^2}{n^3} = -\frac{a^2}{6}+O\left(\frac{1}{n}\right) $$
and also $ \sum_{k=1}^{n}\log\cos\left(\frac{ak}{n\sqrt{n}}\right) \geq -\frac{a^2}{6}+O\left(\frac{1}{n}\right)$. The claim simply follows by exponentiation.
