TITLE: A limit without Taylor series or l'Hôpital's rule $\lim_{n\to\infty}\prod_{k=1}^{n}\cos \frac{k}{n\sqrt{n}}$
QUESTION [7 upvotes]: Computing without Taylor series or l'Hôpital's  rule 
$$\lim_{n\to\infty}\prod_{k=1}^{n}\cos \frac{k}{n\sqrt{n}}$$
What options would I have here? Thanks!

REPLY [4 votes]: The best way (I think) to solve this problem is to use $\ln(1-x)=-x+O(x^2)$ and $\sin x=x+O(x^3)$. In fact,
\begin{eqnarray*}
\lim_{n\to\infty}\ln\prod_{k=1}^n\cos\frac{k}{n\sqrt{n}}&=&\lim_{n\to\infty}\sum_{k=1}^n\ln\cos\frac{k}{n\sqrt{n}}=\lim_{n\to\infty}\frac{1}{2}\sum_{k=1}^n\ln\cos^2\frac{k}{n\sqrt{n}}\\
&=&\lim_{n\to\infty}\frac{1}{2}\sum_{k=1}^n\ln(1-\sin^2\frac{k}{n\sqrt{n}})\\
&=&-\lim_{n\to\infty}\frac{1}{2}\sum_{k=1}^n\left[\left(\frac{k}{n\sqrt{n}}\right)^2+O\left(\left(\frac{k}{n\sqrt{n}}\right)^4\right)\right]\\
&=&-\lim_{n\to\infty}\frac{1}{2}\sum_{k=1}^n\frac{k^2}{n^3}=-\frac{1}{6}.
\end{eqnarray*}
Hence
$$\lim_{n\to\infty}\prod_{k=1}^n\cos\frac{k}{n\sqrt{n}}=e^{-\frac{1}{6}}.$$