$\lim_{ n\to \infty }\sqrt[n]{\prod_{i=1}^n \frac{1}{\cos\frac{1}{i}}}=\,\, ?$ Find the limit :
$$\lim_{ n\to \infty }\sqrt[n]{\prod_{i=1}^n \frac{1}{\cos\frac{1}{i}}}=\,\,?$$
My try :
$$\lim_{ n\to \infty }\sqrt[n]{a} =1\,\, \text {for}  \,\,a>0$$
and;
$$\prod_{i=1}^n \frac{1}{\cos\frac{1}{i}}>0$$
so :
$$\lim_{ n\to \infty }\sqrt[n]{\prod_{i=1}^n \frac{1}{\cos\frac{1}{i}}}=1$$
Is it right?
 A: Let $\epsilon>0$ be given.  Take $N$ so large that $|\log(\cos(1/i))|<\epsilon/2$ whenever $i>N$.  Then, we can write for $n>N+1$
$$\begin{align}
\left|\log\left(\sqrt[n]{\prod_{i=1}^n\sec(1/i)}\right)\right|&=\frac1n\left|\sum_{i=1}^n\log(\cos(1/i))\right|\\\\
&\le\frac1n\sum_{i=1}^N|\log(\cos(1/i))|+\frac1n\sum_{i=N+1}^n|\log(\cos(1/i))|\\\\
&\le \frac1n\sum_{i=1}^N|\log(\cos(1/i))|+\frac\epsilon2 \left(1-\frac{N}{n}\right)\\\\
&<\epsilon
\end{align}$$
when $n>\max\left(N+1,\frac{2N|\log(\cos(1/N))|}{\epsilon}\right)$.
Therefore, we have $\lim_{n\to \infty}\log\left(\sqrt[n]{\prod_{i=1}^n\sec(1/i)}\right)=0$ and

$$\lim_{n\to \infty}\sqrt[n]{\prod_{i=1}^n\sec(1/i)}=1 \tag 1$$



NOTE:  We could have applied the Stolz-Cesaro Theorem and obtained directly

$$\begin{align}
\lim_{n\to \infty}\frac{\sum_{i=1}^n\log(\cos(1/i))}{n}&=\lim_{n\to \infty}\frac{\sum_{i=1}^{n+1}\log(\cos(1/i))-\sum_{i=1}^n\log(\cos(1/i))}{(n+1)-(n)}\\\\
&=\lim_{n\to \infty}\log(\cos(1/(n+1)))\\\\
&=0
\end{align}$$
from which we recover $(1)$ immediately.
A: You got the right answer for the wrong reason, eg. $(2^n)^{1/n} \to 2 \neq 1$.
Cauchy's second limit theorem states
$$\lim_{n \to \infty} (a_n)^{1/n} = \lim_{n \to \infty} \frac{a_{n+1}}{a_n},$$
if the limit on the RHS exists.
This case reduces to 
$$\lim_{n \to \infty} \frac{1}{\cos(1/(n+1))} =1.$$
