If $\sum x_k$ is converges conditionally, then $\sum x^+_k$ diverges

Question

I'm working on this problem

If $\sum x_k$ is a conditionally convergent series in $\mathbb R$, then $\sum x^+_k$ diverges where $x^+_k = \max \{x_k,0\}$.

Could you please verify if my attempt is correct or not? Thank you so much for your help!

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My attempt:

From $\sum |x_k|$ diverges, we get for all $\varepsilon >0$ $$\exists N_0 \in \mathbb N, \forall N>N_0: \sum_{k = 0}^N |x_k| =\sum_{k = 0 \atop x_k \ge 0}^N x_k - \sum_{k = 0 \atop x_k < 0}^N x_k \ge \varepsilon \tag{1}$$

Assume the contrary that $\sum x^+_k$ converges. Then there exists $\varepsilon' >0$ $$\forall N \in \mathbb N: \left | \sum_{k = 0}^N x^+_k \right |= \sum_{k = 0 \atop x_k \ge 0}^N x_k < \varepsilon' \tag{2}$$

From $(1)$ and $(2)$, there exists $\varepsilon' >0$ $$\forall \varepsilon>0,\exists N_0 \in \mathbb N, \forall N>N_0:\varepsilon + \sum_{k = 0 \atop x_k < 0}^N x_k < \varepsilon'$$

As such, $(x_k)_{k \in \mathbb N}$ has a subsequence of negative terms that converges to $ -\infty$. Thus $(x_k)_{k \in \mathbb N}$ diverges, which is a contradiction. In conclusion, $\sum x^+_k$ diverges.

Answer 1

Use the fact that $x_k=x_k^{+}-x_k^{-}$. If $\sum x_k^{+}$ converges then so does $\sum x_k^{-}$ because it is given that $\sum x_k$ converges. It now follows that $\sum |x_k|=\sum x_k^{-} +\sum x_k^{+}$ converges and this is a contradiction.