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Given infinite series $\sum_{n=1}^{\infty} a_n$. Let $a_n^+=max(a_n,0)$, $a_n^-=min(a_n,0)$. Prove that if $\sum_{n=1}^{\infty} a_n$ converges absolutely then $\sum_{n=1}^{\infty} a_n^{+}$ and $\sum_{n=1}^{\infty} a_n^{-} $ converges.

My attempt:

Assume that $\sum_{n=1}^{\infty} a_n^{+}$ and $\sum_{n=1}^{\infty} a_n^{-}$ are the partial sums of $\sum_{n=1}^{\infty} a_n$. Since, $\sum_{n=1}^{\infty} a_n$ converges, their partial sums also converge. Can we make this assumption?

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  • $\begingroup$ Do you mean $\sum_{n=1}^{\infty}a_n$ converges absolutely? Otherwise it's not necessarily true. $\endgroup$ – carmichael561 Apr 18 '16 at 21:27
  • $\begingroup$ yes @carmichael561 $\endgroup$ – combo student Apr 18 '16 at 21:28
  • $\begingroup$ $a_n^\pm \leq |a_n|$ for all $n$, and the result follows by comparison $\endgroup$ – Jon Warneke Apr 18 '16 at 21:32
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Considering that $a_n^{-} \leq 0 \leq a_n^+$, one has

$$a_n^{+}, a_n^{-} \leq a_n^{+}- a_n^{-} = |a_n|$$ Then by the comparison test, you are done.

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