Let me introduce my problem: Let $C \subset \mathbb{C}$ be the unit circle of the complex plane and $Z=\left\{ z_n \right\}_{n \in \mathbb{N}} \subset C$ be a dense subset of the unit circle meaning that $\bar{Z}=C$ or otherwise put that for all $c \in C$ there is a subsequence $\left\{z_{n_{k}}\right\}_{k \in \mathbb{N}}$ of $Z$ such that $z_{n_{k}} \xrightarrow{k} c$. I want to know whether the following series converges:

\begin{align*} \sum_{j\in\mathbb{N}}z_n \end{align*}

For every $z_j$ on the unit circle there is it's anti-diametric $\hat{z_j}=-z_j$ and for all $\varepsilon>0$ we can find an index $\hat{j}\in\mathbb{N}$ so that $|\hat{z_j}-z_{\hat{j}}|<\varepsilon\Leftrightarrow |z_j+z_{\hat{j}}|<\varepsilon$.

If we group all members of the sequence this way (pairwise for a given $\varepsilon$) then the sum can be written in the following form:

\begin{align*} \sum_{j\in\mathbb{N}}z_n=\left(z_1+z_{\hat{1}}\right)+ \left(z_{k_2}+z_{\hat{k_2}}\right)+\ldots \end{align*}

Then by the triangular inequality:

\begin{align*} \sum_{j\in\mathbb{N}}z_n \le |\sum_{j\in\mathbb{N}}z_n| \le \left|z_1+z_{\hat{1}}\right|+ \left|z_{k_2}+z_{\hat{k_2}}\right|+\ldots \le \varepsilon + \varepsilon + \ldots \end{align*}

So this way I don't prove that the series converges. Is there some other way to prove it or it doesn't hold at all? (Anyway I think it was not a good idea to use the absolute value there...).

And a second question on that: If $\left\{a_n\right\}_{n\in \mathbb{N}}$ is a real sequence such that $\sum_{j\in\mathbb{N}}a_n$ converges then $\sum_{j\in\mathbb{N}}a_nz_n$ converges (following the same procedure described above). Can we somehow loosen the conditions on $\left\{a_n\right\}_{n\in \mathbb{N}}$ ? Are there weaker conditions on the sequence $a_n$ so that the series will converge?

Update 1: Does this series converge:

\begin{align*} \sum_{n\in\mathbb{N}}e^{2\pi\vartheta\alpha} \end{align*}

with $\alpha\in \mathbb{R} - \mathbb{Q}$ ?

  • $\begingroup$ Let $z_n = e^{2i\pi \theta n}$, with $\theta$ some irrational number. Then $z_n$ is dense on the unit circle and serves your assumptions. But you can easily pick a series of vectors near to $1$, s.t. the series of partial sums diverges into infinity. Doesn't this contradict your postulate? You can make the series convergent of you just choose any order s.t. suceeding elements cancel out each other sufficiently well. But you may miss elements if you construct that sequence. $\endgroup$
    – shuhalo
    Mar 22, 2011 at 22:13

3 Answers 3


Like others have already said, your series is not convergent because the terms don't get smaller as $n \rightarrow \infty$, and this the case no matter how you order your terms.

What you sketched was instead a proof that if we "pack together" some of the terms and rearrange the packs, then we obtain a series $\Sigma (z_{i_n} + z_{j_n})$ which is convergent.

Indeed, you can pick the index sequences $i_n$ and $j_n$ as follows : Pick $i_n =$ the smallest integer which was not picked for one of the previous $i_k$ or $j_k$ for $k<n$. Then pick $j_n =$ any index such that $|z_{i_n} + z_{j_n}| < 2^{-n}$ which is not already picked.

$j_n$ exists because of the density hypothesis, and with this procedure, you will pick every element of the sequence (the index $k$ will be picked for $n \le k$ somewhere).

So you can obtain a convergent series. However, with some other way of packing things, you will get a divergent series, and you can also get a series that converges to any complex number you want. With this method, your series can do whatever it wants.

There are some kind of series for which no matter how you pack terms together, you will always obtain the same result, they are the absolutely convergent series. If $\Sigma |z_n|$ is convergent, then no matter how you order or pack your terms, you will obtain the same limit everytime. But with other series, anything can happen. You can also turn convergent series who are not absolutely convergent into divergent series if you pack/reorder things in a bad way.

  • 2
    $\begingroup$ The singular of "series" is "series". $\endgroup$
    – joriki
    Mar 24, 2011 at 10:39
  • $\begingroup$ @joriki : so that's why I had a weird feeling when reading everything else ! I guess you're right (but it still seems weird) $\endgroup$
    – mercio
    Mar 24, 2011 at 14:40

This certainly wouldn't be absolutely convergent, so it's convergence is dependent on the order of the terms, yet you do not have any specified order. I can think of an ordering for which the sequence certainly diverges, that is one which starts from the center of the circle and works its way out that looks like: \begin{align*} \sum_{j\in\mathbb{N}}(z_j + \hat{z}_j) \end{align*}


A necessary condition for a series to converge is that the terms go to zero, so no, the series does not converge. However, for some dense sequences in the circle there will be summability methods that apply.

I believe that there also exist real sequences $(a_n)_n$ and sequences $(z_n)_n$ dense in the circle such that $\sum a_n$ converges but $\sum a_n z_n$ does not. To get an example you would have to start with a conditionally convergent series $\sum a_n$ such as the alternating harmonic series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$. Sorry for not making this rigorous, but if you choose a sequence $(z_n)_n$ dense in the circle such that "enough" of the odd terms are "close enough" to $1$ and "enough" of the even terms are "close enough" to $-1$, then you can make the series diverge.


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