Let be $\sum\limits_{k=1}^{n} z_k $ a complex series which converges (only) conditionally.
Show that there exists a bijection $\varphi:\mathbb{N}\to \mathbb{N}$ such that $\lim\limits_{n\to\infty}\left|\sum\limits_{k=1}^{n} z_{\varphi(k)}\right| =\infty$.
My (edited) approach:
For each $k\in\mathbb{N}$ let be $z_k=a_k i + b_k$. If $\lim\limits_{n\to\infty}\left|\sum\limits_{k=1}^{n} b_k\right|=\infty$ or $\lim\limits_{n\to\infty}\left|\sum\limits_{k=1}^{n} a_k\right|=\infty$ then $$ \left|\sum\limits_{k=1}^{n} a_k i + b_k\right|=\sqrt{\left(\sum\limits_{k=1}^{n} a_k\right)^2+\left(\sum\limits_{k=1}^{n} b_k\right)^2}\geq \left|\sum\limits_{k=1}^{n} a_k\right|\left|\sum\limits_{k=1}^{n} b_k \right|$$ which would imply that $\sum\limits_{k=1}^{n} z_k$ doesn't converge (Cauchy-Schwarz inequality). If $\sum\limits_{k=1}^{n} a_k$ oscillates and $\sum\limits_{k=1}^{n} b_k $ converges and then I will always find a small enough $\epsilon >0$ such that for each index $n_0$ I find a $n$ with $n>n_0$ and $\sum\limits_{k=n_0+1}^{n} a_k>\epsilon$. Hence, $$ \left|\sum\limits_{k=n_0+1}^{n} a_k i + b_k\right|=\sqrt{\left(\sum\limits_{k=n_0+1}^{n} a_k\right)^2+\left(\sum\limits_{k=n_0+1}^{n} b_k\right)^2}\geq \left|\sum\limits_{k=n_0+1}^{n} a_k\right|>\epsilon, $$ which again would imply that $\sum\limits_{k=1}^{n} z_k$ doesn't converge. In both cases we get a contradiction, hence both series must converge.
Further, if both limits $\sum\limits_{k=1}^{\infty} |b_k|$ and $\sum\limits_{k=1}^{\infty} |a_k|$ exist at the same time then it follows: $$ \sum\limits_{k=1}^{n} \left|a_k i + b_k\right|\leq\sum\limits_{k=1}^{n} |a_k i |+\sum\limits_{k=1}^{n} |b_k |= \sum\limits_{k=1}^{n} |a_k|+\sum\limits_{k=1}^{n} |b_k|. $$ This would mean that $\sum\limits_{k=1}^{n} z_k $ converges unconditionally which again is a contradiction. So at least one series must diverge (towards $\infty$).
WLOG let be $\sum\limits_{k=1}^{\infty} |b_k|$ the divergent (towards $\infty$) series. We define two sets: $I_-$ which contains all the indices of terms $b_k<0$ and $I_+$ which contains all indices of the terms $b_k\geq 0$. Now we consider two real series which are made by indices of $I_+$ and indices of $I_-$ respectively, $\sum\limits_{k=1}^{n} b_k^+$ and $\sum\limits_{k=1}^{n} b_k^-$. If only one of the series diverges then $\sum\limits_{k=1}^{n} b_k$ diverges which is a contradiction. If both converges then it follows that $\sum\limits_{k=1}^{n} |b_k|$ converges which is also a contradiction. Hence, both series $\sum\limits_{k=1}^{n} b_k^+$ and $\sum\limits_{k=1}^{n} b_k^-$ must diverge (towards $\infty$). Let be $\epsilon >0$ arbitrarily chosen. We define a bijection $\varphi:\mathbb{N}\to\mathbb{N}= I_+\cup I_-$ (note that $I_+$ and $I_-$ are clearly a disjoint decomposition of $\mathbb{N}$): $$\begin{align*} &\varphi(1)=\min\{k\in I_-\}\\ &\varphi(2)=\min\{k\in I_+\}\\ &\varphi(3)=\min\{k\in I_+\setminus \{\varphi(2)\}\}\\ &\vdots\\ &\varphi(n_0)=\min\{k\in I_+\setminus \{\varphi(2), \varphi(3), \cdots , \varphi(n_0-1)\}\},\\ \end{align*} $$ where we increment $n_0$ until reach $$ \left|\sum\limits_{k=1}^{n_0}z_{\varphi(k)}\right|= \left|\sum\limits_{k=1}^{n_0} a_{\varphi(k)} i + b_{\varphi(k)}\right|=\sqrt{\left(\sum\limits_{k=1}^{n_0} a_{\varphi(k)}\right)^2+\left(\sum\limits_{k=1}^{n_0} b_{\varphi(k)}\right)^2}\geq \left|\sum\limits_{k=1}^{n_0} b_{\varphi(k)} \right|>\epsilon. $$ Then we start the procedure again: $$\begin{align*} &\varphi(n_0+1)=\min\{k\in I_-\setminus \{\varphi(1)\}\}\\ &\varphi(n_0+2)=\min\{k\in I_+\setminus \{\varphi(2), \varphi(3), \cdots , \varphi(n_0)\}\}\\ &\varphi(n_0+3)=\min\{k\in I_+\setminus \{\varphi(2), \varphi(3), \cdots , \varphi(n_0)\}, \varphi(n_0+2)\}\\ &\vdots\\ &\varphi(n_1)=\min\{k\in I_+\setminus \{\varphi(2), \varphi(3), \cdots , \varphi(n_0),\varphi(n_0+2),\cdots , \varphi(n_1-1)\}\},\\ \end{align*} $$ until we reach $$ \begin{split} \left|\sum\limits_{k=n_0+1}^{n_1}z_{\varphi(k)}\right| &= \left|\sum\limits_{k=n_0+1}^{n_1} a_{\varphi(k)} i + b_{\varphi(k)}\right|\\ & =\sqrt{\left(\sum\limits_{k=n_0+1}^{n_1} a_{\varphi(k)}\right)^2+\left(\sum\limits_{k=n_0+1}^{n_1} b_{\varphi(k)}\right)^2}\geq \left|\sum\limits_{k=n_0+1}^{n_1} b_{\varphi(k)} \right|>\epsilon. \end{split} $$ The function $\varphi$ is injective because every element of each both sets $I_+$ and $I_-$ is chosen once. If there existed an element $i_-\in I_-$ which would have no preimage under $\varphi$ then by construction of $\varphi$ it must be possible to add infinitely many $i_+\in I_+$ such that $\left|\sum\limits_{k=1}^{n} b_{\varphi(k)} \right|$ will always remain below $\epsilon$. However, this means that $\sum\limits_{k=1}^{n} b_k^+$ converges which is a contradiction. (The same argument can be used if we assume that there existed an $i_+\in I_+$ which has no preimage). To summarize, we have found a bijection $\varphi$ which rearranges the series $\sum\limits_{k=1}^{n} z_{\varphi(k)}$ such that we can push $\left|\sum\limits_{k=1}^{n} z_{\varphi(k)}\right|$ above any value $M$.
Is this correct so far?