Prove that if $\sqrt[n]{\prod\limits_{i\leq n}a_i}$ converges to a finite limit then $a_n$ converges

Question

Prove that if $$\lim_{n\to\infty} \sqrt[n]{\prod_{i\leq n}a_i} < \infty$$ then $\lim_{n\to\infty} a_n$ exists.

Given that $\{a_i\}$ is bounded and positive.

So I used Cesaro means to show that $\lim_{n\to\infty} \sqrt[n]{\prod_{i\leq n}a_i} = \lim_{n\to\infty} a_n$ if the limit exists, but how to prove that the limit exists?

Answer 1

The existence of $\lim_{n\to\infty}\Big[\prod_{i\leq n}a_i\Big]^{\frac{1}{n}}$ does not imply the existence of $\lim_{n\to\infty}a_n$, even if we assume that all the $a_n$ are positive.

For instance, if we let $a_n=2$ if $n$ is odd and $a_n=\frac{1}{2}$ if $n$ is even, then $(a_1\cdots a_n)^{\frac{1}{n}}\to 1$ as $n\to\infty$, but $\lim_{n\to\infty}a_n$ does not exist.

Essentially $(a_1\cdots a_n)^{\frac{1}{n}}$ makes the sequence more regular in a manner very similar to Cesaro means, as can be seen by taking logarithms.