0
$\begingroup$

I have been given the following theorem: If the sequence $\{a_n\}$ is convergent, then $\{a_n\}$ is bounded.

This theorem comes with a proof, which I got a couple of questions about, which hopefully somebody here can help me out with. The proof goes like this:

We know that $\{a_n\} \rightarrow l$, for some $l \in \mathbb{R}$. Thus, $\{a_n\}-l$ is a null sequence, so there is a integer $N$ such that

$$|a_n-l| \lt 1$$

for all $n \gt N$. Now,

$$|a_n| = |(a_n-l)+l| \leq |(a_n-l)| +|l| \Leftrightarrow$$

$$|a_n| \lt 1 +|l|$$

Now, as the next step, the book goes:

To complete the proof, we put $K=|a_1| + |a_2| + \cdots + |a_N| + 1 + |l|$, and it then follows that

$$|a_n| \leq K$$

Ok, what happened there? Why does adding the sum of $N$ terms add the possibility of it being equal to K?

$\endgroup$
  • $\begingroup$ Surely you mean the sequences is bounded and not unbounded? $\endgroup$ – Alex R. Feb 10 '15 at 18:28
  • $\begingroup$ What are you doing with your $K$??? If you have, for $n>N$, $|a_n|<1+|l|$, then you have always $|a_n|<K$, $\forall K>1+|l|$.... $\endgroup$ – Martigan Feb 10 '15 at 18:29
  • $\begingroup$ I have a feeling that by "unbounded" you really mean "bounded." $\endgroup$ – Matt Samuel Feb 10 '15 at 18:29
  • $\begingroup$ samuel+alex, yes, . You are absolutely correct. $\endgroup$ – JustDanyul Feb 10 '15 at 18:30
  • $\begingroup$ I don't know why this proof is so complicated, I can think of a two line proof. $\endgroup$ – Matt Samuel Feb 10 '15 at 18:33
1
$\begingroup$

Using the triangle inequality : $$|(a_n-l) + l| \leq |a_n-l| + |l| \leq 1 + |l|$$ because $|a_n-l| \leq 1$ ($a < b \implies a + c < b + c$)

Your $K$ is the sum of the terms for which $|a_k-l|$ could be over 1, hence their sum is clearly over a single of them. The inequality could be strict here but it's not even needed (you don't need the inequality to be strict for the sequence to be bounded). You should always keep a weak inequality when you don't need it to be strict (in case there is some strange case you would have forgotten for example all 0 or something) !

I would rather take $\max(|a_1|, ..., |a_N|, |l|+1)$ (and in this case they could be equal so a weak inequality is needed) but the sum will also work indeed.

$\endgroup$
  • $\begingroup$ thanks a lot for your answer. I suppose it doesn't make much difference, since we are changing from strong to weak. I just thought there was some reason for the change, but I'm guessing not. Again, thanks a lot :-) $\endgroup$ – JustDanyul Feb 10 '15 at 18:52
  • $\begingroup$ I think the reason is you know that weak inequality is trivial while strong inequality could sometimes be wrong in some strange cases so when you are writing a proof where strong is not needed, you'd better go with weak. $\endgroup$ – servabat Feb 10 '15 at 18:56

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.