Show that if a series is conditionally convergent, then the series obtained from its positive terms is divergent, and the series obtained from its negative terms is divergent.
Please, help me to verify or identify flows in my proof:
Let $\sum x_n$ be convergent but $\sum |x_n|$ be divergent i.e. $\sum x_n$ is conditionally convergent.Since the series is not absolutely convergent then by ex.$(9.1.1)$ it has infinitely many negative terms. Let $s_m$ be partial sums of our original series.
Let $t_m$ be partial sums associated with only positive terms. Let's show that $(t_m)$ it is not bounded above. Assume it is bounded above then there is some $K\in \mathbb N$ such that $|t_m|=t_m < K$ for any $m \in \mathbb N$ by $(3.7.5)$Let $q_m$ be the partial sums of the negative terms. Note that for any $m \in \mathbb N$ we have $s_m \le t_m < K$ Moreover, note that $s_m - t_m = q_m$. Hence, $s_m - K < q_m$ i.e. $s_m - q_m < K$. Since $(s_m)$ is convergent than it is bounded by some $L \in \mathbb N$ i.e. $|s_m| \le L$. Hence, we have $|q_m| = |s_m - t_m| \le |s_m| + |t_m| < L + K$
Since we have infinitely many negative terms then there is some $H \in N$ such that if $m > H$ then $|q_m| > L+K$ i.e. a contradiction. Hence, $t_m$ is not bounded for all $m \in \mathbb N$. Hence, $\sum_{x_n > 0} x_n$ is divergent.
If $(q_m)$ is convergent then $(t_m)$ is also convergent since $(s_m)$ is convergent and $t_m = s_m - q_m$. However, we already showed that $(t_m)$ is divergent. Hence, $(q_m)$ is divergent.