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?

  • 1
    $\begingroup$ Try to prove for $k=2$ and then use the induction. $\endgroup$ 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.