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?
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
$$\lim_{n \to \infty} \frac{1}{\cos(1/(n+1))} =1.$$
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.
Much less elegant than RRL'answer.
Consider $$P_n=\left(\prod _{i=1}^n \sec \left(\frac{1}{i}\right)\right){}^{\frac{1}{n}}$$ $$\log(P_n)=\frac 1n \sum _{i=1}^n \log\left(\sec \left(\frac{1}{i}\right)\right)$$ Now, by Taylor for large values of $i$ $$\sec \left(\frac{1}{i}\right)=1+\frac{1}{2 i^2}+\frac{5}{24 i^4}+O\left(\frac{1}{i^6}\right)$$ $$\log\left(\sec \left(\frac{1}{i}\right)\right)=\frac{1}{2 i^2}+\frac{1}{12 i^4}+O\left(\frac{1}{i^6}\right)$$ $$\sum _{i=1}^n \log\left(\sec \left(\frac{1}{i}\right)\right)=\frac{H_n^{(2)}}{2}+\frac{H_n^{(4)}}{12}+\cdots$$ where appears harmonic numbers.
Using asymptotics $$\log(P_n)=\frac 1 n \left(\frac{90 \pi ^2+\pi ^4}{1080}-\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)\right)$$ Taylor again $$P_n=e^{\log(P_n)}=1+\frac{90 \pi ^2+\pi ^4}{1080 n}+O\left(\frac{1}{n^2}\right)$$ which seems to be a quite reasonable approximation $$\left( \begin{array}{cccc} n & P_n & \text{approximation} & \text{difference} \\ 10 & 1.09393 & 1.09127 & 0.00266 \\ 20 & 1.04713 & 1.04563 & 0.00149 \\ 30 & 1.03145 & 1.03042 & 0.00103 \\ 40 & 1.02360 & 1.02282 & 0.00078 \\ 50 & 1.01889 & 1.01825 & 0.00063\\ 60 & 1.01574 & 1.01521 & 0.00053 \\ 70 & 1.01349 & 1.01304 & 0.00046 \\ 80 & 1.01181 & 1.01141 & 0.00040 \\ 90 & 1.01050 & 1.01014 & 0.00036 \\ 100 & 1.00945 & 1.00913 & 0.00032 \end{array} \right)$$