# Limit $\mathop {\lim }\limits_{n \to \infty } \frac{{n{{\left( {{a_1}...{a_n}} \right)}^{\frac{1}{n}}}}}{{{a_1} + ... + {a_n}}}$

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.

• 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 $1$.
– 5xum
Feb 24, 2014 at 14:48
• You right, I didn't mention $a_i>0.\forall i$. Feb 24, 2014 at 14:49
• By AM-GM the limit is smaller equal than $1$.
– J.R.
Feb 24, 2014 at 14:50
• I know that if $a_n$ is a constant, then the limit equals $1$. What about the other case? I understood that it might be equal to $2\over e$. How exactly? Feb 24, 2014 at 14:56
• 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 $(a_n)_n)$
– J.R.
Feb 24, 2014 at 15:01

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 $$0$$.