Let $a_n \in \mathbb{R}$, such that $\sum_{n=1}^\infty|a_n|= \infty$ and $\sum_{n=1}^m a_n \to a$, as $m \to \infty$. Let $a_n^+=\max\{a_n,0\}.$ Show that $\sum_{n=1}^\infty a_n^+= \infty$.
Approach:
We know, in case of real numbers, $\max\{a,b\}=\displaystyle\frac{1}{2}(a+b+|a-b|)$. Hence, $a_n^+=\max\{a_n,0\}=\displaystyle\frac{1}{2}(a_n+|a_n|)$. Therefore, the sum becomes,
$\displaystyle\sum_{n=1}^\infty a_n^+=\sum_{n=1}^\infty \displaystyle\frac{1}{2}(a_n+|a_n|)=\frac{1}{2}a+\frac{1}{2}\sum_{n=1}^\infty|a_n|=\infty$
As far as I can remember, I asked this question here earlier (but I have forgotten entirely how I tackled it; I don't want to steal ideas from the past). But this time, I am asking this again here in order to get this verified (not much confident with my procedure).
Thank you!