Prove that if $\sum_{n=1}^{\infty} a_n$ converges then $\sum_{n=1}^{\infty} a_n^{+}$ and $\sum_{n=1}^{\infty} a_n^{-}$ converges.

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?

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

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.