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?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Pedro
    Jun 1, 2021 at 19:13

3 Answers 3


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 $|a+ib| \ge |a|$.


Some comments:

  1. The notation $\sum_{k=1}^\infty z_k = \infty$ is not standard. Here it's also rather ambiguous: do you mean you want it to converge to positive real infinity with zero imaginary component? Or perhaps $\lim_{n \to \infty} \left|\sum_{k=1}^n z_{\varphi(k)} \right| = \infty$?

  2. "Both limits $\sum\limits_{k=1}^{\infty} b_k$ and $\sum\limits_{k=1}^{\infty} a_k$ must exist."

    Yes, because taking the real and imaginary parts are continuous operations.

  3. "Otherwise it follows that $\left|\sum\limits_{k=1}^{n} a_k i + b_k\right|=\sqrt{\left(\sum\limits_{k=1}^{n} a_k\right)^2+\left(\sum\limits_{k=1}^{n} b_k\right)^2}\geq \left|\sum\limits_{k=1}^{n} b_k \right|$ which would imply that $\sum\limits_{k=1}^{\infty} z_k$ doesn't converge (same would follow if you choose $\sum\limits_{k=1}^{n} |a_k|$ as a lower bound). This is a contradiction."

    This does not make sense. You could for instance abstractly have $b_k$ oscillating between $-1$ and $1$ which would make your bound pointless.

  4. A general point: you repeatedly say a thing, give some partial reasons why the thing is true, then reiterate the thing shortly thereafter. It gets perhaps a little wordy or tiresome, especially when the sub-arguments are so short. For instance, rather than an outline like, "Both series must diverge. Because if only one diverges [...]. And if both converge [...]. Thus both series diverge." you could do, "If both converged [...], so at least one diverges. If exactly one diverges, [...], so both diverge."

  5. where we increment $n_0$ until reach $\sum\limits_{k=1}^{n_0}z_{\varphi(k)}>\epsilon$

    You need to define $\epsilon$ before you use it. Here you may as well just use $1$ instead. Since $z$ is complex, your inequality makes no sense. Maybe you meant $b_{\varphi(k)}$.

  6. I am not sure how to show that $\varphi$ is surjective.

    I mean, it's pretty clear from your construction that you'll keep using a single value from $I_-$ followed by some positive number of values from $I_+$ [ok, your literal construction could perhaps use 0 values from $I_+$ in a given stage, but just don't do that], which will eventually exhaust them. Formally you'd appeal to something like the axiom of dependent choice or induction. Doesn't seem worthwhile to me.

  • $\begingroup$ You like to have $n$ as the upper limit of summation in the first comment when taking the limit as $n\to\infty$. $\endgroup$
    – Gary
    Jun 1, 2021 at 8:42
  • $\begingroup$ @Gary Typo fixed. $\endgroup$ Jun 1, 2021 at 10:22
  • $\begingroup$ Thank you for your great tips :) I have edited my approach and incorporated your suggestions. However, I am not sure if I am allowed (as a 2nd year math undergrad) to argue as you proposed. So I have provided some more arguments. Hopefully it is correct now. $\endgroup$
    – Philipp
    Jun 1, 2021 at 12:20

I was looking at the argument in the answer, and then I noticed it had been edited. It appears that the edit may be similar to what I wrote below, but I will post it anyway since I think it may be a bit simpler, and hopefully, easier to understand.

This is a part of the Riemann Rearrangement Theorem.

One or Both of the Real or Imaginary Part Does Not Converge Absolutely

  1. Since $\sum\limits_{k=1}^\infty z_k=\sum\limits_{k=1}^\infty(x_k+iy_k)$ converges, both $\sum\limits_{k=1}^\infty x_k$ and $\sum\limits_{k=1}^\infty y_k$ must converge.

  2. If both $\sum\limits_{k=1}^\infty|x_k|$ and $\sum\limits_{k=1}^\infty|y_k|$ converge, then $\sum\limits_{k=1}^\infty|z_k|\le\sum\limits_{k=1}^\infty|x_k|+\sum\limits_{k=1}^\infty|y_k|$ converges.

  3. Since $\sum\limits_{k=1}^\infty|z_k|$ diverges, at least one of $\sum\limits_{k=1}^\infty|x_k|$ or $\sum\limits_{k=1}^\infty|y_k|$ must diverge.

Without loss of generality assume that $\sum\limits_{k=1}^\infty|x_k|$ diverges.

Partition the Indices

Partition $\mathbb{N}$ into $\{p_k\}$ and $\{n_k\}$ so that $$p_{k+1}\gt p_k\quad\text{and}\quad n_{k+1}\gt n_k$$ and $$x_{p_k}\ge0\quad\text{and}\quad x_{n_k}\lt0$$ That is, $\{p_k\}$ are the indices of the non-negative terms and $\{n_k\}$ are the indices of the negative terms.

The Sum of Both Subsequences Diverge

  1. If both $\sum\limits_{k=1}^\infty x_{p_k}$ and $\sum\limits_{k=1}^\infty x_{n_k}$ converge, then, since they converge absolutely, $\sum\limits_{k=1}^\infty\left|x_k\right|=\sum\limits_{k=1}^\infty x_{p_k}-\sum\limits_{k=1}^\infty x_{n_k}$ also converges.

  2. If one of $\sum\limits_{k=1}^\infty x_{p_k}$ or $\sum\limits_{k=1}^\infty x_{n_k}$ converges and the other diverges, then $\sum\limits_{k=1}^\infty x_k=\sum\limits_{k=1}^\infty x_{p_k}+\sum\limits_{k=1}^\infty x_{n_k}$ diverges.

  3. Since $\sum\limits_{k=1}^\infty x_k$ converges and $\sum\limits_{k=1}^\infty\left|x_k\right|$ diverges, both $\sum\limits_{k=1}^\infty x_{p_k}$ and $\sum\limits_{k=1}^\infty x_{n_k}$ must diverge.

Compute $\boldsymbol{\varphi}$

Carry out the following pseudo-code to compute $\varphi$:

$\quad\mathsf{while}(s\lt m)\{\varphi(k)=p_i;s=s+x_{p_i};\,i=i+1;k=k+1\};m=m-1;$
$\quad\mathsf{while}(s\gt m)\{\varphi(k)=n_j;s=s+x_{n_j};j=j+1;k=k+1\};m=m+2;$

Since both $\sum\limits_{k=1}^\infty x_{p_k}$ and $\sum\limits_{k=1}^\infty x_{n_k}$ diverge, both of the $\mathsf{while}$ loops will terminate and repeat endlessly.

Since $p_k$ and $n_k$ are a partition of $\mathbb{N}$, $\varphi$ is bijective.

Since $s\ge m$ at the end of the first $\mathsf{while}$ loop, we have $$ \sum_{k=1}^{\psi(m)}x_{\varphi(k)}\ge m $$ which ensures the divergence of $$ \sum_{k=1}^\infty x_{\varphi(k)} $$


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