What I am thinking so far is following:

Construct another sequence $\{b_n\}$ such that $b_1=x_2, b_2=x_3, \ldots, b_n=x_{n+1}.$ Since $\{x_n\}$ is a Cauchy sequence, $\{b_n\}$, as a subsequence of $\{x_n\},$ is therefore also a Cauchy sequence.

Since Cauchy sequences are bounded, we can find $M_1, M_2 \in Q$ such that $|x_n| \le M_1$ and $|b_n| \le M_2$ for all $n.$ Let $M = \max\{M_1, M_2\}.$

Since $\{x_n\}$ is a Cauchy sequence, we can find $N_1$ such that for all $m,n \ge N_1,$ $|x_n-x_m| < \frac \varepsilon{2M}$. Similarly, we can find $N_2$ such that for all $m,n \ge N_2,$ $|b_n-b_m| < \frac \varepsilon{2M}.$

Let $N = \max\{N_1, N_2\}$ and $n,m \ge N.$

Therefore, $$|y_n-y_m|=|x_n x_{n+1}-x_m x_{m+1}|=|x_n b_n-x_m b_m|=|x_n b_n-x_m b_n+x_m b_n-x_m b_m|=|b_n||x_n-x_m|+|x_m||b_n-b_m| \le M\times\frac \varepsilon{2M} + M\times\frac \varepsilon{2M} = \varepsilon$$


I wonder if there is any flaw in this proof and if there is a better way to prove it. Thank you!

  • 1
    $\begingroup$ Please, use MathJax to format mathematical expressions in your posts. $\endgroup$ – mucciolo Dec 10 '17 at 2:45
  • $\begingroup$ The idea looks right to me! One minor mistake: Near the end, in the line "Therefore...", when you use the triangle inequality to break up an absolute value, you should have $\leq$ instead of $=$: specifically, $|x_nb_n-x_mb_n+x_mb_n-x_mb_m| \leq |x_nb_n-x_mb_n| + |x_mb_n-x_mb_m|$. But this doesn't affect your argument. As far as better way: I can't think of anything. You could simplify notation slightly (just write $x_{n+1}$ instead of $b_n$, etc.) but nothing significant. $\endgroup$ – Zach Teitler Dec 10 '17 at 3:19
  • $\begingroup$ No need to introduce the sequence $\{b_n\}$ just to obtain the bound for $\{x_{n+1}\}$. If $|x_n| \le M$ for all $n \in \mathbb{N}$ then also $|x_{n+1}| \le M$ for all $n \in \mathbb{N}$. Great otherwise. $\endgroup$ – mechanodroid Dec 10 '17 at 9:58

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