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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.

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    $\begingroup$ What do you know about the values of $a_i$? For example, if only one of them equals $0$, the limit is $0$, if all of them value $1$, then the limit will be $1$. $\endgroup$ – 5xum Feb 24 '14 at 14:48
  • $\begingroup$ You right, I didn't mention $a_i>0.\forall i$. $\endgroup$ – AndrePoole Feb 24 '14 at 14:49
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    $\begingroup$ By AM-GM the limit is smaller equal than $1$. $\endgroup$ – J.R. Feb 24 '14 at 14:50
  • $\begingroup$ @TooOldForMath, nice! thanks $\endgroup$ – AndrePoole Feb 24 '14 at 14:54
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    $\begingroup$ The proof of AM-GM is nothing but the inequality $1+x\le e^x$, so you're essentially looking at the $\frac{1+x}{e^x}$, no make $x$ really large and this is not going to be a good estimate, i.e. the exact value of the limit (or whether it even exists?) depends on the $(a_n)_n)$ $\endgroup$ – J.R. Feb 24 '14 at 15:01
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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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