Geometric mean of a positive convergent sequence Let $(x_n)$ be a positive sequence such that $x_n \to x $. It is required to show that $(x_1x_2...x_n)^{\frac{1}{n}}\to x$. All the proofs I know of this require the AM-GM inequality. Is there any way to show this using the basic properties of convergent sequences?(no logs allowed)

Use the "cut the sequence and analyse the main part and the tail separately".
Steps:


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*Pick an $\epsilon > 0$ and then there is an $N$ such that $|x_n - x| < \epsilon$ for any $n \geq N$. So, $x-\epsilon \leq x_n \leq x + \epsilon$. (Warning $x - \epsilon$ might be negative... how do you deal with that?)

*For $m \geq N$, cut the product $x_1 \cdots x_m$ into tow halfs. The second half ($x_i$, where $i \geq N$) you deal with using $1$. Now, using Step 1, try to find a $N_1$ such that if $n \geq N_1$ then both parts in steps $1$ and $2$.
This is a VERY standard trick in analysis, which you should get used to. This is the same proof, basically, that you would use to show that $\frac{x_1 + \cdots + x_n}{n} \to x$.
Note: It is often easier conceptually (and practically) to try to normalize the thing you are dealing with. However, in this case it adds more steps needlessly, but it is easier to understand. E.g. if $x > 0$, then you can replace the original sequence $x_n$, with $y_n = x_n/x$ so that $y_n \to 1$ and you now want to show that $\sqrt[n]{y_1\cdots y_n} \to 1$, because multiplying both sides by $x$ gives the original result. For $x = 0$ (because the limit of positive numbers might be zero), this boils down to a proof no easier than the original proof: eventually $x_n$ is small.
