# Prove that if $\lim_{n\to\infty} x_n = 1$ for $x_n > 0$ then $\lim_{n\to\infty} \sqrt[n]{x_1x_2\cdots x_n} = 1$

Given a sequence $$x_n$$ and the fact that: $$\lim_{n\to\infty} x_n = 1\\ x_n > 0\\ n\in\Bbb N$$ Prove $$\lim_{n\to\infty} \sqrt[n]{x_1x_2\cdots x_n} = 1$$

I'm having some difficulties finishing this proof. I've shown while solving another problem that: $$\lim_{n\to\infty} x_n = a \implies \lim_{n\to\infty}{1\over n}\sum_{k=1}^nx_k = a$$

Using this we may state that: $$\lim_{n\to\infty}x_n = 1 \implies \lim_{n\to\infty}{1\over n}\sum_{k=1}^nx_k = 1$$

On the other hand by AM-GM we have that: $$\frac{x_1 + x_2 + \cdots x_n}{n} \ge \sqrt[n]{x_1x_2\cdots x_n}$$

Since $$x_n > 0$$: $$\frac{x_1 + x_2 + \cdots x_n}{n} \ge \sqrt[n]{x_1x_2\cdots x_n} \ge 0$$

We know that: $$\lim_{n\to\infty}\frac{x_1 + x_2 + \cdots x_n}{n} = 1$$

Therefore: $$1 \ge \lim_{n\to\infty}\sqrt[n]{x_1x_2\cdots x_n} \ge 0$$

My idea was to use Monotone Convergence theorem, but since $$x_n$$ is only constrained by $$x_n > 0$$ we can not make any conclusions on the monotonicity of: $$y_n = \sqrt[n]{x_1x_2\cdots x_n}$$ (or can we?).

Apparently my idea to use MCT is not applicable here. So the question is what would be the proper way to prove the above?

• Can't we say that the average of the logarithms goes to $0?$ Feb 28, 2019 at 17:44
• @metamorphy Proving Stolz-Cesaro follows right after this problem. So i may not use it yet. Feb 28, 2019 at 17:45
• But you said you've proved $x_n\to a\implies (1/n)\sum x_k \to a$ Isn't that enough? We know $\log{x_n}\to 0.$ Feb 28, 2019 at 17:51
• @saulspatz you are right, guess I need a rest Feb 28, 2019 at 17:52

Consider the logarithm and use Stolz cesaro to deduce that $$\lim_{n\to \infty}\log\sqrt[n]{x_1\dotsb x_n}=\lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^n\log x_i=\lim_{n\to \infty}\log x_n=0$$ since $$x_n\to 1$$ from which the claim follows.

• Thank you, i was missing such an obvious thing Feb 28, 2019 at 17:58

Consider the sequence defined, for all $$n \geq 0$$, by$$y_n = \ln(x_n)$$

Obviously $$y_n \rightarrow 0$$. So by the Cesaro theorem, you have that $$\frac{y_1 + ... + y_n}{n} \rightarrow 0$$

That means that $$\frac{\ln(x_1 \times ... \times x_n)}{n} \rightarrow 0$$

i.e. $$\ln \left( \sqrt[n]{x_1 \times ... \times x_n} \right) \rightarrow 0$$

You deduce that $$\sqrt[n]{x_1 \times ... \times x_n} \rightarrow 1$$

Just for the completeness: we can use the $$HM\le GM\le AM$$ inequality: $$H_n:=\frac{1}{\frac{\sum_{k=1}^n {\frac{1}{x_k}}}{n}} \le \left(\prod_{k=1}^n {x_k}\right)^{\frac{1}{n}}\le \frac{\sum_{k=1}^n {x_k}}{n}=:A_n$$

• $$x_n\to 1 \text{ with } x_n>0 \implies \frac{1}{x_n}\to 1$$
• Cesaro summation Cauchy's first limit theorem implies that $$\frac{1}{H_n}\to 1$$ (so $$H_n\to 1$$ ) and $$A_n\to 1$$
• apply the Squeeze theorem