Here is the question:

Let $b_n$ be a sequence of positive numbers such that $b_n \rightarrow 0$. Suppose $|a_n - a_m| \leq b_n$ for all $m \geq n$. Prove that $a_n$ is Cauchy.

Attempt at a solution: Since $b_n \rightarrow 0$ and $b_n > 0$ for all $n$, we know that $b_n < \epsilon$ for given $\epsilon > 0$. Since $|a_n - a_m| \leq b_n$ for all $m \geq n$, $|a_n - a_m| < \epsilon$ for all $m \geq n$. Let $N \leq n$. Then $N \leq m$, so for all $m \geq n \geq N$, $|a_n - a_m| < \epsilon$, which fulfills the definition of a Cauchy sequence.

I'm not entirely sure if this proof is correct, though I am unsure of where (if anywhere) I went wrong.

  • 2
    $\begingroup$ What comes after "Since..." is, at least, very inaccurate. It should be "since $\;b_n\to0\;$ , for any $\;\epsilon>0\;$ there exists $\;N\in\Bbb N\;$ s.t. $\;n>N\implies b_n<\epsilon\;$ ." (did you see where I used that $\;b_n>0\;$ ?) You can then continue as you wrote... $\endgroup$
    – DonAntonio
    Jan 30 '18 at 17:34
  • 1
    $\begingroup$ Yes you have used the positivity by replacing $|b_n|<\epsilon$ by $b_n<\epsilon$ $\endgroup$ Jan 30 '18 at 17:35
  • 1
    $\begingroup$ The idea is correct, to tighten this up, I suggest you, write down the definition of the Cauchy criterion, and then fit your work to show that the definition has been satisfied. $\endgroup$
    – Doug M
    Jan 30 '18 at 17:44

Since $b_n\to 0$ and $b_n>0$ for all $n$, we know that $b_n<\epsilon$ for given $\epsilon >0$.

You've suddenly moved from using $n$ as an unrestricted value ("for all $n$") to using it in ways that are evidently not universal ($b_n<\epsilon$) without properly stating so. Also, the most obvious interpretation (for someone who doesn't know how to do the proof themselves and is therefore trying to learn it from your proof) would be that you mean a fixed value of $n$, dependent only on $\epsilon$.

Since $|a_n−a_m|≤b_n$ for all $m\ge n, |a_n−a_m|<\epsilon$ for all $m\ge n$.

Okay. So now we know that for any $m$ greater than that fixed $n, |a_n−a_m|<\epsilon$.

Let $N\le n$. Then $N\le m$, so for all $m \ge n \ge N, |a_n−a_m|<\epsilon$, which fulfills the definition of a Cauchy sequence.

Really? I can just pick a value $N \le n$ and this works?. Well, I pick $N = 0$. Now we have two ways of interpreting the next sentence:

  • For all $m$, but still with that same fixed $n$, when $m \ge n \ge N, |a_n-a_m| < \epsilon$. This is what you showed before. but it does not fulfill the definition of a Cauchy sequence, because that definition requires $n$ to vary as well.
  • For all $m$, and for all $n$ with $m\ge n\ge N$. But now, you have no longer shown $|a_m - a_n| < \epsilon$. That was only shown to be true for the fixed value of $n$ chosen earlier. Even worse, since $N$ was arbitrarily chosen, you are claiming it is true even for lower values of $n$ than the one you picked originally. So you have not met the Cauchy condition here, either.

$N$ is the linch-pin of the Cauchy condition. It must be picked with care to satisfy its purpose. It should be picked based on $\epsilon$ right at the start.

Fortunately $b_n \to 0$ means more than $b_n < \epsilon$ for some $n$. The definition of convergence here (using that $b_n > 0$ for all $n$) is

"Given $\epsilon > 0$, there is an $N$ such that for all $n \ge N, b_n < \epsilon$".

Now, $N$ is the fixed value, and $n$ still varies freely over its restricted domain.


Your Answer

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