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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 $\lim_{n \to \infty} \sqrt[2n]{a_{2n+1}} = \lim_{n \to \infty} \sqrt[n]{a_n} = 1$ ?
william_grisaitis
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    Note that $\sqrt[2n]{a_1a_2\cdots a_{2n+1}}=\exp\left(\frac{2n+1}{2n}\log\sqrt[2n+1]{a_1a_2\cdots a_{2n+1}}\right)$. – user1551 Jul 31 '23 at 18:43
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    Alternatively, $\frac{1}{2}a < a_{2n + 1} < 2a$ for sufficiently large $n$. – Gary Jul 31 '23 at 23:47

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