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?
Best Answer
I don't have enough time to study your argument in detail; I hope Joshua P. Swanson's answer addressed your problems here. However, I would recommend adding some structure to your argument by dividing the proof in smaller pieces.
For example, are you familiar with the real case of your problem? If yes, the complex case is easily reduced (see Step 2). If not, see Step 1 and try to tackle it separately, without the notational noise coming from the more general case.
Step 1. If $\sum_{k=1}^\infty x_k$ is a real conditionally convergent series, then after rearranging $\sum_{k=1}^\infty x_{\varphi(k)} = +\infty$.
This is a special case of the Riemann rearrangement theorem. The construction is the following: first start the rearrangement with as many positive $x_k$'s as is needed to obtain a sum larger than $1$. Second, add one nonpositive $x_k$. Next, add some positive terms to obtain a sum larger than $2$, add one nonpositive term, and so on.
Step 2. The complex case.
Denote the real and imaginary part: $z_k = a_k + i b_k$. Since $\sum z_k$ converges, both $\sum a_k$ and $\sum b_k$ converge as well. If also both $\sum |a_k|$ and $\sum |b_k|$ converged, the inequality $|z_k| \le |a_k|+|b_k|$ would imply that $\sum |z_k|$ convergences too - and we know this is false. So we may assume (without loss of generality) that $\sum a_k$ converges only conditionally.
Taking the rearrangement as in Step 1 (for $x_k = a_k$), we obtain $$ \left|\sum_{k=1}^N z_k\right| = \left|\sum_{k=1}^N a_k + i \sum_{k=1}^N b_k\right| \ge \left|\sum_{k=1}^N a_k\right| \xrightarrow{N \to \infty} \infty, $$ thanks to the inequality $|a+ib| \ge |a|$.