0
$\begingroup$

If $ \sum_{1}^{\infty} a_n $ be absolutely convergent then how to show $ \sum_{1}^{\infty} a_n $ is also convergent?

The proof of this question in my textbook:

Let $ s_k := \sum_{1}^{k} a_n $ and $ t_k := \sum_{1}^{k} |a_n|$.

Then using m>k, $$ |s_m - s_k|= |\sum_{k+1}^{m}a_n| \leq \sum_{k+1}^{m}|a_n| = t_m -t_k \to 0. $$


I am really unsure about the above inequality and how that has been used to prove the theorem ? Also I understand that $|t_m -t_k| < \epsilon$, but I am unsure how this implies that $t_m-t_k \to 0$

$\endgroup$
  • $\begingroup$ The inequality is called triangle inequality. $\endgroup$ – Santosh Linkha Dec 17 '16 at 13:18
  • $\begingroup$ The only inequality in that proof is one that follows from the triangle inequality: what isn't clear here? $\endgroup$ – DonAntonio Dec 17 '16 at 13:18
  • $\begingroup$ @DonAntonio I am not sure why that inequality is true ? A proof would be much appreciated $\endgroup$ – questiontime Dec 17 '16 at 13:24
  • $\begingroup$ @iheartalgebraa I think that someone dealing with analysis should long time ago have studied that. Anyway, if you google "triangle inequality" I'm sure you'll find millions of sites... $\endgroup$ – DonAntonio Dec 17 '16 at 13:27
  • 1
    $\begingroup$ @iheartalgebraa You are not using it "in the case of series". You are using it "in the case of finite sums". $\endgroup$ – user228113 Dec 17 '16 at 13:35
1
$\begingroup$

The proof of this question in my textbook:

let $ s_k := \sum_{1}^{k} a_n $ and $ t_k := \sum_{1}^{k} |a_n|$. Then using $m>k$, $$ \big|s_m - s_k\big|= \biggr|\sum_{k+1}^{m}a_n\biggr| \leq \sum_{k+1}^{m}|a_n| = t_m -t_k \to 0 .\tag{1}$$

I am really unsure about the above inequality

Well, try to prove it using $|a+b|\leq|a|+|b|$.

and how that has been used to prove the theorem ?

Keep the goal in mind: you want to show that $$ |s_m-s_k|<\epsilon $$ for large enough $m$ and $k$.

Also I understand that |t_m -t_k| < e but I am unsure how this implies that t_m- t_k -> 0

$t_m-t_k\to 0$ is a bad writing. What it really means here is $|t_m-t_k|<\epsilon$ for "large enough" $m$ and $k$ where $\epsilon>0$ is assumed to be given.


Essentially the inequality in (1) tells you that $\{s_k\}$ is a Cauchy sequence.

$\endgroup$
0
$\begingroup$

Use Cauchy's criterion: the series is convergent if and only if for any $\varepsilon>0$, there exists a $N>0$ such that for any $m,k\ge N$, we have $\;\lvert s_m-s_k\rvert <\varepsilon$.

Now since the series is absolutely convergent, we do have a $N$ such that for any $m,k\ge N$, $\;\lvert t_m-t_k\rvert <\varepsilon$. By the triangle inequality, the same is a fortiori true for $\lvert s_m-s_k\rvert$.

Sketch of the proof of Cauchy's criterion:

It's the same criterion for sequences and for series. So let $(u_n)$ be a Cauchy sequence. Proving it has a limit goes through the following steps:

  • We associate to $(u_n)$ two sequences: $(a_n)=(\inf\limits_{p\ge n}u_p)$ and $(b_n)=(\sup\limits_{p\ge n}u_p)$.
  • Show $(a_n)$ is non-decreasing and $(b_n)$ is non-increasing. Furthermore, for all $n$, we have $$a_n\le u_n\le b_n.$$
  • Deduce from Cauchy's property that $(b_n-a_n)$ tends to $0$, so they're adjacent sequences and converge to a common limit $L$. By the squeezing principle, $(u_n)$ also converges to $L$.
$\endgroup$
  • 1
    $\begingroup$ I think you missed the point of the OP: he already has this proof. He just seems to be lost on the triangle inequaity. $\endgroup$ – DonAntonio Dec 17 '16 at 13:29
  • $\begingroup$ @bernard I understand that |t_m -t_k| < e but I am unsure how this implies that t_m-t_k -> 0 $\endgroup$ – questiontime Dec 17 '16 at 13:32
  • $\begingroup$ @iheartalgebraa: It is not that $t_m-t_k\to 0$ (how is it defined anyway? it's not a term of a sequence, as there are 2 indices), but that it can be made as small as you please, choosing $m,k$both large enough. This is simply Cauchy's criterion. $\endgroup$ – Bernard Dec 17 '16 at 13:37
  • $\begingroup$ @Bernard can you expand on explaining the proof of cauchy's criterion because this is what is limiting my understanding of this proof. $\endgroup$ – questiontime Dec 17 '16 at 13:39
  • $\begingroup$ I've added some details.Does that make it clearer to you? $\endgroup$ – Bernard Dec 17 '16 at 14:32
0
$\begingroup$

Since $\sum a_n$ is absolutely convergent, so $\sum |a_n|$ is convergent. And since $a_n\le|a_n|$, by comparison test, $\sum a_n$ is convergent.

$\endgroup$
  • 4
    $\begingroup$ You can't use comparison test here as it is not given $\;a_n\;$ is positive.., in fact, if it is a positive series then the claim is very boring. The theorem is interesting only for non-positive series. $\endgroup$ – DonAntonio Dec 17 '16 at 13:25

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.