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Suppose $x_{n}>0$ for all $n\in {\bf N}$ and $\lim_n x_{n}=x_{0}>0$. How can I show that $$ \lim_n \left( x_{1}x_{2}...x_{n}\right)^{\frac {1} {n}}=x_{0}\quad ? $$

furkans
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2 Answers2

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We have $$ \log \left( ( x_1 x_2 ... x_n)^{1/n} \right) = \frac{1}{n} \log(x_1 x_2 ... x_n) = \frac{1}{n} ( \log(x_1) + \log(x_2) + ... + \log(x_n) ) = \frac{\log(x_1) + \log(x_2) + ... + \log(x_n)}{n} $$

As $ \lim_{n \rightarrow \infty } \log(x_n) = \log( \lim_{ n \rightarrow \infty } x_n ) = \log(x_0) $, we deduce that $$ \lim_{n \rightarrow \infty } \frac{\log(x_1) + \log(x_2) + ... + \log(x_n)}{n} = \lim_{ n \rightarrow \infty } \log(x_n) = \log(x_0) $$ And finally $$ \lim_{ n \rightarrow \infty } \log \left( ( x_1 x_2 ... x_n)^{1/n} \right) =\log \left( \lim_{ n \rightarrow \infty } ( x_1 x_2 ... x_n)^{1/n} \right) = \log(x_0) \longrightarrow \lim_{ n \rightarrow \infty } ( x_1 x_2 ... x_n)^{1/n} = x_0 $$

M159
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  • You should show how you deduced that $$\lim_{n \to \infty } \frac{\log(x_1) + \log(x_2) + ... + \log(x_n)}{n} = \lim_{ n \to \infty } \log(x_n) = \log(x_0)$$While it is true, the readers who are unfamiliar with, for example, the Stolz-Cesaro Theorem, might not understand why it is true. – Mark Viola Jul 10 '17 at 16:13
  • But the user who asked this has this question: https://math.stackexchange.com/q/2352889/9464 – M159 Jul 11 '17 at 08:33
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Let $f(n)$ be defined by

$$f(n)=\left(\prod_{m=1}^n x_m\right)^{1/n}$$

Clearly the logarithm of $f(n)$ is given by

$$\log(f(n))=\frac1n \sum_{m=1}^n\log(x_m)$$

From the Stolz-Cesaro Theorem, we have

$$\begin{align} \lim_{n\to \infty}\log(f(n)) &=\lim_{n\to \infty} \left(\frac1n \sum_{m=1}^n\log(x_m)\right)\\\\ &=\lim_{n\to \infty}\left( \frac{\sum_{m=1}^{n+1}\log(x_m)-\sum_{m=1}^{n}\log(x_m)}{(n+1)-n}\right)\\\\ &=\lim_{n\to \infty}\log(x_{n+1})\\\\ &=\log(x_0) \end{align}$$

Mark Viola
  • 179,405