Vtyjkyuk

Let $(x_n)$ be a positive sequence such that $x_n to x $. It is required to show that $(x_1x_2...x_n)^frac1nto x$. 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:

1. 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?)




2. 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 $fracx_1 + cdots + x_nn to x$.

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 $sqrt[n]y_1cdots y_n to 1$, 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.

Another approach to avoid AM-GM:
$$beginalign
lim_ntoinftyln (x_1x_2cdot x_n)^1/n
&=lim_ntoinftyfracln x_1+ln x_2+cdots+ln x_nn \
&=lim_ntoinftyfrac(ln x_1+cdots+ln x_n+1)-(ln x_1+cdots+ln x_n)(n+1)-n qquad(1)\
&=lim_ntoinftyln x_n+1\
&=ln x
endalign
$$

Thus we can conclude
$$colorredlim_ntoinfty(x_1x_2cdots x_n)^1/n=x$$

$(1)$: Stoltz-Casearo theorem is used.