# Product of Cauchy Sequences is Cauchy

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?

• Try to prove for $k=2$ and then use the induction. Oct 19 at 23:54

$$\newcommand{\seq}[1]{\left( #1_k\right)_{k \in \Bbb N}} \newcommand{\e}{\varepsilon} \newcommand{\N}{\mathbb{N}} \newcommand{\abs}[1]{\left| #1 \right|}$$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.

• 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? Oct 20 at 0:22
• Sure, I'm pretty sure you can do that, if I understand you correctly. Oct 20 at 0:33
• Thanks a lot for the answer! It really helped! Oct 20 at 0:39