$ \lim_{n\to \infty} (a_1a_2...a_n)^{1/n} = a$ and we know that $ \lim_{n\to \infty} a_n = a$
My proof: $| (a_1a_2...a_n)^{1/n} - a| <| (a_na_n...a_n)^{1/n} - a| = | (a_n)^{n/n} - a| = | a_n - a| < \epsilon $
My question is: is this proof correct? If no, where did I do a mistake?
Being that you are calculating the n-root I would assume that $a_n>0$
$$\lim_{n\to \infty} a_n = a \iff \lim_{n\to \infty} \ln{a_n} = \ln{a}$$
Then:
$$\lim_{n\to \infty} (a_1a_2...a_n)^{1/n}=\exp{\ln{\lim_{n\to \infty} (a_1a_2...a_n)^{1/n}}}=\exp{\lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^n\ln{a_i}}$$
Using Cesaro Means:
$$\exp{\lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^n\ln{a_i}}=\exp{\ln{a}}=a$$
