Can I claim that $\sum _n ^{\infty }a _n=\sum _n ^{\infty }a _n^++\sum _n ^{\infty }a _n^- $ How can I motivate the following.
I'm writing a proof for that if a seris $\sum a _n $ converges conditionally then the series consisting of its negative terms and the series consisting of its positive terms must both diverge. And my proof hinge of the following:
I separete the positive terms from the negative by writing $a _n ^+=(a _n + |a _n |)/2 $ and $a _n ^- = (a _n - |a _n |)/2 $
Then how can I motivate that I can write
$$\sum _n ^{\infty }a _n=\sum _n ^{\infty }a _n^++\sum _n ^{\infty }a _n^-   $$
Since from here I can argue that both series on the right must diverge, but I'm not sure I'm correct to claim the equality between the both sides.
Thanks in advance!
 A: You can write $$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty (a_n^++a_n^-),$$
but writing what you did seems a bit of a stretch. You can argue, however, that if both "half sums" converge, then the full sum will also converge.
I do not see how that means both series must diverge.
A: If a seris $\sum a _n $ converges absolutelly then the series consisting of its negative terms and the series consisting of its positive terms must both diverge.
Since $a _n ^+=\frac{(a _n + |a _n |)}{2} \le \frac{(|a _n| + |a _n |)}{2}=|a_n|$, then $\sum _n ^{\infty }a _n^+ $ converges.
It is easy to see that $\sum _n ^{\infty }a _n^- $ also converges.
Therefore $$\sum _n ^{\infty }a _n=\sum _n ^{\infty }a _n^++\sum _n ^{\infty }a _n^-   $$
A: Let $\displaystyle\sum_{n=1}^{\infty}a_{n}$ be a conditionally convergent series, so $\displaystyle\sum_{n=1}^{\infty}a_{n}$ converges but $\displaystyle\sum_{n=1}^{\infty}\left|a_{n}\right|$ diverges.
Since the term-by-term sum or difference of two series is divergent 
if one series converges and the other diverges, it follows that
$\displaystyle\sum_{n=1}^{\infty} a_{n}^{+}=\frac{1}{2}\sum_{n=1}^{\infty}\left(a_n+\left|a_n\right|\right)$  $\;$ and $\;\;$$\displaystyle\sum_{n=1}^{\infty} a_{n}^{-}=\frac{1}{2}\sum_{n=1}^{\infty}\left(a_n-\left|a_n\right|\right)$ $\;\;$ both diverge.
A: To be rigorous, it might be easier to do things via partial sums. 
Let $S_n = \sum_{m=1}^n a_m$, let $T_n = \sum_{m=1}^n a_m^+$, and let $U_n = \sum_{m=1}^n a_m^-$.
The statement that $\sum_{n=1}^{\infty} a_n$ converges is equivalent to the statement that 
$\lim_{n \rightarrow \infty} S_n$ exists and is finite. If $\sum_{n=1}^{\infty} a_n^+$ also
converges, then $\lim_{n \rightarrow \infty} T_n$ exists and is finite, so $\lim_{n \rightarrow \infty} (2T_n - S_n)$ exists and is finite. 
But $2a_n^+ - a_n = 2a_n^+ - a_n^+ + a_n^- = a_n^+ + a_n^- = |a_n|$. Hence $2T_n - S_n = \sum_{m=1}^n |a_m|$.
So the statement that $\lim_{n \rightarrow \infty} (2T_n - S_n)$ exists and is finite means that the series converges absolutely. Hence if both $\sum_{n=1}^{\infty} a_n$ and 
$\sum_{n=1}^{\infty} a_n^+$ converge, the series converges absolutely.
A similar argument can be used if If $\sum_{n=1}^{\infty} a_n^-$ converges. So if the series converges conditionally, neither $\sum_{n=1}^{\infty} a_n^+$ nor $\sum_{n=1}^{\infty} a_n^-$ can converge.... since in either case the result would be that your series converges absolutely, which doesn't occur under conditional convergence.
