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I want know an example of real sequence $x_n\gt0$ which satisfies

$\frac1n (x_1+x_2+...+x_n)$diverges but $(x_1x_2...x_n)^{\frac1n}$ converges.

I know if $x_n$ converges so that $\frac1n (x_1+x_2+...+x_n)$ and $(x_1x_2...x_n)^{\frac1n}$

and if $x_n$ is increasing sequence whose limt is infinite than $lim\frac1n (x_1+x_2+...+x_n)$ and $lim(x_1x_2...x_n)^{\frac1n}$ also infinite.

So i thought if $x_n$ has two subseuence which converges in different value than it can be the example but i couldn't find it.

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up vote 5 down vote accepted

Hint: Try the sequence $1, 1, 2,\frac{1}{2}, 3,\frac{1}{3}, 4,\frac{1}{4}, 5,\frac{1}{5}\dots$. The divergence of the sequence $\frac{1}{n}(x_1+\cdots +x_n)$ should be straightforward. For the convergence of $(x_1x_2\cdots x_n)^{1/n}$, you will need information about the behaviour of something closely related to $n^{1/n}$.

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Thanks. it's very helpful. – Hwan May 5 '14 at 15:17
You are welcome. There are a number of ways to modify the example. For instance, leave the $1,2,3,4,\dots$ alone but make the "small" entries go to $0$ very fast. – André Nicolas May 5 '14 at 15:26

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