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Prove that for every sequence $(a_n)_n$ of complex numbers, if the series $\sum_{n\ge 0} a_n$ is absolutely convergent, then $|\sum_{n\ge 0} a_n| \le \sum_{n \ge 0} |a_n|$.

I've been given the following hint as well which I guess has to be used:

You may use the fact that if $(z_n)_n$ is a sequence of complex numbers and $\displaystyle\lim_{n ->\infty} (z_n)$ exists, then the limit of the real sequence $(|z_n|)_n$ exists and $\displaystyle\lim_{n->\infty}|z_n| = \displaystyle\lim_{n->\infty} (|z_n|)$

I don't really understand the hint... They're just saying in the final statement that X=X are they not?

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  • $\begingroup$ There seems to be a typo. The hint should be: $|\displaystyle\lim_{n\to\infty} z_n|= \displaystyle\lim_{n\to\infty} |z_n| $. $\endgroup$ – Braindead Dec 11 '14 at 15:05
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Welcome to MSE.

Judging from the problem it seems like what you are being asked to do is actually two things:

  1. Show that $\displaystyle\sum_{n\ge0} a_n$ is convergent.
  2. Show that $\left|\displaystyle\sum_{n\ge0} a_n\right| \le \displaystyle\sum_{n\ge0} |a_n|$

To show the first part, you should show that the sequence of partial sums

$s_k = \displaystyle\sum_{n=0}^k a_n$ is a Cauchy sequence. This can be done using the Triangle inequality and the fact that $t_k=\displaystyle\sum_{n=0}^k |a_n|$ is a Cauchy sequence. (Why is $t_k$ a Cauchy sequence?)

Once you have established convergence of of $\lim s_n$, you just need to use Triangle inequality again and the fact that limits pass through inequalities, along with the corrected hint, and you are done.

Let me know if you need further help.

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  • $\begingroup$ A typo: you mean $\le$ and not equality in (2)... $\endgroup$ – user1537366 Dec 11 '14 at 15:26
  • $\begingroup$ @user1537366 Thanks, the typo was fixed. $\endgroup$ – Braindead Dec 11 '14 at 15:29
  • $\begingroup$ @Braindead Thanks a lot for your help. We aren't given any actual series, so I guess we are meant to prove the inequality for a general $a_n$ based on the initial condition. $\endgroup$ – Douglas Dec 11 '14 at 15:35
  • $\begingroup$ @Douglas That's right, but the first thing you should do is to prove that $\sum a_k$ converges based on the fact that $\sum |a_k|$ converges. $\endgroup$ – Braindead Dec 11 '14 at 15:40
  • $\begingroup$ Sorry, I'm not quite understanding, what is $a_k$? and how would we prove the sum of $a_k$ converges based on the fact the modulus of it does..? $\endgroup$ – Douglas Dec 11 '14 at 15:50

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