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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?

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

3 Answers 3

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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.

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  • $\begingroup$ Thank you, i was missing such an obvious thing $\endgroup$
    – roman
    Feb 28, 2019 at 17:58
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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 $$ \sqrt[n]{x_1 \times ... \times x_n} \rightarrow 1$$

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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
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