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

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.

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.

• 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? Commented Dec 6, 2013 at 18:08
• @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... Commented Dec 6, 2013 at 18:11
• 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 $\sum a_n$ Commented Dec 6, 2013 at 18:22
• then, lets say the two limits are $L,M$ then the limit of $\sum|a_n|$ is $|L|+|M|$ Commented Dec 6, 2013 at 18:24
• @DanielGagnon That's right! Commented Dec 6, 2013 at 22:25