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I'm trying to disprove the following statement: If $(s_{n,1})$,$(s_{n,2})$,...,$(s_{n,k})$ are Cauchy sequences, then the sequence $\prod_{i=1}^{k} s_{n,i}$ is also Cauchy.

I'm not too sure what I'm supposed to do here. How do I prove that it's true, or how do I come up with a counterexample? What is the intuition behind the product of Cauchy sequences?

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    $\begingroup$ Try to prove for $k=2$ and then use the induction. $\endgroup$ Oct 19 at 23:54
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$ \newcommand{\seq}[1]{\left( #1_k\right)_{k \in \Bbb N}} \newcommand{\e}{\varepsilon} \newcommand{\N}{\mathbb{N}} \newcommand{\abs}[1]{\left| #1 \right|} $A good tip in math when faced with problems like these is to try small examples, and to see what that gets you.

For instance, try $k=2$, $k=3$, and so on; maybe the process can be generalized. See what steps you have to make time and time again.


Let's take $k=2$ and prove that.

So, for notational simplicity, let $\seq a, \seq b$ be Cauchy. Take the sequence $\seq c$ defined by $c_k := a_k b_k$. Since you've tagged this with "real analysis", I will also assume you're working in $\Bbb R$ and hence will assume the sequences are bounded by $M$.

Let $\e > 0$. We wish to show $\exists N \in \N$ such that, for all $m,n \ge N$,

$$\abs{c_m - c_n} = \abs{a_m b_m - a_n b_n} < \e$$

Begin by adding and subtracting the term $a_m b_n$:

\begin{align*} \abs{a_m b_m - a_n b_n} &= \abs{a_m b_m + a_m b_n - a_m b_n + a_n b_n }\\ &\le \abs{ a_m b_m - a_m b_n } + \abs{ a_m b_n - a_n b_n } \\ & = \abs{a_m} \abs{b_m - b_n} + \abs{b_n} \abs{ a_m - a_n } \\ & \le M \abs{b_m - b_n} + M \abs{ a_m - a_n } \end{align*}

Since $\seq a, \seq b$ are Cauchy, then $\exists N_1,N_2 \in \N$ such that, whenever $n,m \ge N$,

$$\abs{a_m - a_n} < \frac{\e}{2M} \qquad \abs{b_m -b_n} < \frac{\e}{2M}$$

Therefore, if we take $N := \max(N_1,N_2)$ and let $n,m \ge N$,

$$\abs{a_m b_m - a_n b_n} \le M \abs{b_m - b_n} + M \abs{ a_m - a_n } < M \frac{\e}{2M} + M \frac{\e}{2M} = \e$$

yielding that $\seq c$ is Cauchy.


Now try this for $k=3$. See what patterns emerge.

Alternatively, you can make the simpler argument that since $\seq a,\seq b$ Cauchy implies $\seq{a_k b}$ Cauchy, then - more clearly - the product of any two Cauchy sequences is Cauchy.

So if we take $\seq a, \seq b, \seq c$, known to be Cauchy, then the product of $\seq{a_k b}$ and $\seq c$ is Cauchy, i.e. the sequence $\seq{a_k b_k c}$ is Cauch (since each individually are)y. And so on and so forth.

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  • $\begingroup$ I see. If we were to use Induction in this manner, can I redefine the sequence with terms that are the product of the first k Cauchy sequences as a new sequence itself? $\endgroup$ Oct 20 at 0:22
  • $\begingroup$ Sure, I'm pretty sure you can do that, if I understand you correctly. $\endgroup$ Oct 20 at 0:33
  • $\begingroup$ Thanks a lot for the answer! It really helped! $\endgroup$ Oct 20 at 0:39

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