I had this question in mind when I was revisiting my earlier question: Getting stuck on an Analysis Question - Limit Theory . Now that I know $\lim\limits_{n\to\infty} \sqrt[n]{X_1\cdot X_2\cdot \cdot \cdot X_n} = a$ if $\lim\limits_{n\to\infty}X_n = a$ and $X_n > 0, a>0$. I want to see if it is okay to prove this proposition if $a=0$.

My intuitive idea is this:

For all positive $\epsilon$, there is some $N$, if $n \geq N, |x_n|< \epsilon$. Thus, somehow (though I don't know how to work out the details), $\lim\limits_{n\to\infty} \sqrt[n]{X_N\cdot X_{N+1}\cdot \cdot \cdot X_n} < \lim\limits_{n\to\infty} \epsilon^{\frac{n-N+1}{n}} = \epsilon$

And the previous terms ${X_1\cdot X_2\cdot \cdot \cdot X_{N-1}}$ are real. Hence, we can say $\lim\limits_{n\to\infty} \sqrt[n]{X_1\cdot X_2\cdot \cdot \cdot X_{N-1}} = 1$. Thus, establishing specific $N_1$ and $N_2$ respectively, can we conclude that $\lim\limits_{n\to\infty} \sqrt[n]{X_1\cdot X_2\cdot \cdot \cdot X_n} = 0 < \epsilon$ ?

(I wish I could figure this out myself but I failed. Thanks for your help!)

  • $\begingroup$ Overall it looks good. But I don't know why you want to establish $N_1$ and $N_2$. At that point one usually concludes that $\lim\limits_{n\to\infty} \sqrt[n]{X_1\cdot X_2\cdot \cdot \cdot X_n} < \epsilon$ for all positive $\epsilon$. This implies $\lim\limits_{n\to\infty} \sqrt[n]{X_1\cdot X_2\cdot \cdot \cdot X_n} = 0 $. $\endgroup$
    – user103093
    Feb 28, 2016 at 8:15

1 Answer 1


Your idea is great. To complete the answer following your idea, let $\epsilon >0$. Then there is $N$ so that $|x_n|<\epsilon$ whenver $n\ge N$. Since we have

$$\lim_{n\to \infty} \sqrt[n]{X_1X_2\cdots X_{N-1}} =1 \text{ and } \lim_{n\to \infty} \sqrt[n]{\epsilon^{-N+1}} =1,$$

there is $N_1 \ge N$ so that

$$\sqrt[n]{X_1X_2\cdots X_{N-1}} \le 2, \text{ and } \sqrt[n]{\epsilon^{-N+1}} \le 2 \ \ \forall n\ge N_1. $$ Then

$$\begin{split} \sqrt[n]{X_1X_2\cdots X_n} &= \sqrt[n]{X_1X_2\cdots X_{N-1}} \sqrt[n]{X_N X_{N+1} \cdots X_n} \\ &\le 2 \epsilon^{\frac{n-N+1}{n}}\\ & \le 4\epsilon \end{split}$$ whenever $n\ge N_1$. Since $\epsilon$ is arbitrary, this implies

$$\lim_{n\to \infty} \sqrt[n]{X_1X_2\cdots X_n} =0.$$

  • $\begingroup$ I realize this way doesn't work because the part $\sqrt[n]{\epsilon^{-N+1}} \le 2 \ \ \forall n\ge N_1.$ is infeasible as we can always find an extremely small $\epsilon$ which breaks the inequality. $\endgroup$
    – Davista
    Mar 5, 2016 at 5:38
  • 1
    $\begingroup$ @Davista . Yes, but note that we first fix $\epsilon$, then choose our $N_1$ large. Of course if $\epsilon$ gets smaller, one needs to find a bigger $N_1$. $\endgroup$
    – user99914
    Mar 5, 2016 at 5:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .