I'm aware that $\lim_{n \to \infty} \sqrt[n]{a_1 a_2 \dotsm a_n} = a$, as shown by If $(x_n) \to x$ then $(\sqrt[n]{x_1x_2\cdots x_n}) \to x$.
My question is, what is $\lim_{n \to \infty} \sqrt[2n]{a_1 a_2 \dotsm a_{2n+1} }$?
Assuming $\lim_{n \to \infty} a_n = a > 0$, I think we have
$$ \displaystyle \begin{align} \lim_{n \to \infty} \sqrt[2n]{a_1 a_2 \dotsm a_{2n+1} } & = \lim_{n \to \infty} \sqrt[2n]{a_1 a_2 \dotsm a_{2n} } \sqrt[2n]{a_{2n+1}} \\ & = a \lim_{n \to \infty} \sqrt[2n]{a_{2n+1}} \end{align} $$
My question then becomes:
- what is $\lim_{n \to \infty} \sqrt[2n]{a_{2n+1}} $?
- does $\lim_{n \to \infty} \sqrt[2n]{a_{2n+1}} $ exist?
- if $\lim_{n \to \infty} a_n > 0$, is $\lim_{n \to \infty} \sqrt[2n]{a_{2n+1}} $ simply $1$, because $\lim_{n \to \infty} \sqrt[2n]{a_{2n+1}} = \lim_{n \to \infty} \sqrt[n]{a_n} = 1$ ?