Prove: $$\sum {{a_{{n_k}}}} < \infty \Rightarrow \sum {|{a_n}| < \infty } $$

In words, if every sub-series of $\sum a_n$ converges then $\sum a_n$ converges absolutely.
I know that:
$$\liminf\sum {{a_n}} \le \sum {{a_{{n_{}}}}} \le \limsup \sum {{a_n}} $$ therefore, $\sum a_n$ is bounded, which implies $\sum a_n$ converges, but it doesn't tell me if the series converges absolutely.

Other direction crossed my mind is splitting the series into two sub-series; odd-indices-series and even-indices-series, but I can't see how it can be helpful in this case.


1 Answer 1


Hint: by assumption, the subseries consisting of positive terms converges, and the subseries consisting of negative terms also. Use this to show that the series converges absolutely.

  • $\begingroup$ So, you're saying: $\sum {|{a_n}} | \le max(\liminf\sum {|{a_n}} |,\lim \sup \sum {|{a_n}|} ) < \infty $. therefore, $\sum |a_n|$ converges. Is that right? $\endgroup$ Commented Dec 6, 2013 at 18:08
  • $\begingroup$ @DanielGagnon No, that's not what I'm saying! What I'm saying is that the partial sums of $\sum |a_n|$ are the difference of the partial sums of the two subseries that I described. If these two subseries converge, then... $\endgroup$ Commented Dec 6, 2013 at 18:11
  • $\begingroup$ Oh, I think I get it know. You splitted the series into two subseries: one consists the positive terms, and the other consists all the negative terms. So, the difference is a finite number which is $\sum a_n$ $\endgroup$ Commented Dec 6, 2013 at 18:22
  • $\begingroup$ then, lets say the two limits are $L,M$ then the limit of $\sum|a_n|$ is $|L|+|M|$ $\endgroup$ Commented Dec 6, 2013 at 18:24
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    $\begingroup$ @DanielGagnon That's right! $\endgroup$ Commented Dec 6, 2013 at 22:25

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