A limit without Taylor series or l'Hôpital's rule $\lim_{n\to\infty}\prod_{k=1}^{n}\cos \frac{k}{n\sqrt{n}}$ 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!
 A: The following tries to use only "basic" inequalities, avoiding L'Hopital and Taylor. 
The inequality 
$$\tag1 e^x\ge 1+x\qquad x\in\mathbb R$$
should be well-known and after taking logarithms immediately leads to
$$\tag2 \ln(1+x)\le x\qquad x>-1$$
and after taking reciprocals
$$\tag3 e^{-x}\le \frac1{1+x}\qquad x>-1 $$
Substituting $-\frac x{1+x}$ for $x$ in $(3)$ and taking logarithms, we have
$$\tag4 \frac x{1+x}\le \ln(1+x)\qquad x>-1.$$
For $0<x<\frac\pi2$ we have 
$$\ln\cos x=\ln\sqrt{1-\sin^2x}=\frac12\ln(1-\sin^2x)$$
and by $(2)$ and $(4)$
$$\tag 5 -\frac12\tan^2x=\frac12 \frac{-\sin^2x}{1-\sin^2x}\le \ln\cos x\le -\frac12\sin^2x.$$
Since $0\le \sin x \le x\le \tan x$ for $0\le x<\frac\pi2$ and the quotient of the upper and the lower bound in $(5)$ is just $\cos^2x$ we find for $0<x\le y<\frac\pi2$ (using $\cos^2y\le \cos^2x$)
$$-\frac12 x^2\cdot\frac1{\cos^2y}\le \ln\cos x\le -\frac12 x^2\cdot \cos^2y.$$
Letting (for $n\ge1$) $y=\frac1{\sqrt n}$, $x=\frac k{n\sqrt n}$ with $1\le k\le n$, this gives
$$-\frac{k^2}{2n^3}\cdot\frac{1}{\cos^2\frac1{\sqrt n}}\le \ln\cos \frac k{n\sqrt n}\le -\frac{k^2}{2n^3}\cdot\cos^2\frac1{\sqrt n}.$$
Using $1^2+2^2+\ldots + n^2=\frac{n(n+1)(2n+1)}{6}$ we find by summation
$$ -\frac{n(n+1)(2n+1)}{12n^3}\cdot\frac{1}{\cos^2\frac1{\sqrt n}}\le\sum_{k=1}^n\ln\cos\frac k{n\sqrt n}\le-\frac{n(n+1)(2n+1)}{12n^3}\cdot\cos^2\frac1{\sqrt n}$$
and by taking the limit
$$ \sum_{k=1}^\infty\ln\cos\frac k{n\sqrt n}=-\frac16$$
and ultimately
$$ \prod_{k=1}^\infty\cos\frac k{n\sqrt n}=e^{-\frac16}.$$
A: 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
$$\lim_{n\to\infty}\prod_{k=1}^n\cos\frac{k}{n\sqrt{n}}=e^{-\frac{1}{6}}.$$
A: I am just going to prove that the product converges.
To get the exact value,
 you need more advanced stuff.
Start with $|\cos(x)-1| \le x^2/2$.
This can be proved from
$\cos(2x) = \cos^2(x)-\sin^2(x) = 1 - 2 \sin^2(x)$,
so $|\cos(2x)-1| = 2|\sin^2(x)|$.
Since $|\sin(x)| \le x$ for
all $x$,
$|\cos(2x)-1| \le 2x^2$
or $|\cos(x)-1| \le x^2/2$.
Since $|\cos(x)| \le 1 + x^2/2$,
$|\cos(\frac{k}{n\sqrt{n}})|
\le 1 + \frac{k^2}{2n^3}
$.
Since
$\sum_{k=1}^n \frac{k^2}{2n^3}
= \frac1{2n^3} \sum_{k=1}^n k^2
= \frac{n(n+1)(2n+1)}{12n^3}
< 1
$
for $n > 1$,
the product converges
(since $\prod (1+a_n) < \exp(\sum a_n)$ 
if $a_n > 0$).
