Evaluate the following limit.
$a_i > 0.\forall i\in \mathbb{N}$.
$$\mathop {\lim }\limits_{n \to \infty } \frac{{n{{\left( {{a_1}...{a_n}} \right)}^{\frac{1}{n}}}}}{{{a_1} + ... + {a_n}}}$$
I tried to use the Stolz-Cesaro Theorem which I thought might feet here. I got this expression:
$$\mathop {\lim }\limits_{n \to \infty } \frac{{(n + 1){{\left( {{a_1}...{a_{n + 1}}} \right)}^{\frac{1}{{n + 1}}}} - n{{\left( {{a_1}...{a_n}} \right)}^{\frac{1}{n}}}}}{{{a_{n + 1}}}}$$
Which looked a little promising, but I don't know how to take it from here.
Answer
I believe the limit can be anything between $0$ and $1$.
Taking $a_n=1$ for all $n$ clearly yields the limit of $1$.
On the other hand, if you take $$a_n=\cases{1&$\text{ if }$ n$\neq 2^k$\\ \frac1n&\text{ else}},$$
You can prove that the limit of $$\frac{a_1+\dots+a_n}{n}$$ is still $1$ (it would be $1$ even if you replaced every $2^k$-th value with $0$), while the limit of $$\sqrt[n]{a_1a_2\cdots a_n} = 0,$$
meaning the total limit is $0$.
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