A subseries of the series $\displaystyle\sum _{n=1}^\infty a_n$ is defined to be a series of the form $\displaystyle\sum _{k=1}^\infty a_{n_k}$, for $n_k \subseteq \Bbb N$.

Prove or disprove: For any conditionally convergent series $\displaystyle\sum _{n=1}^\infty a_n,\ \exists\ k\geq 2\ $ such that the subseries $\displaystyle\sum _{n=1}^\infty a_{kn}$ converges.

Certainly, every conditionally convergent series has some subseries that converge (decreasing alternating terms), and some subseries that diverge (the subseries of all of the positive terms; also the subseries of all of the negative terms).


My original "answer" which I now doubt is a counter-example to the proposition:

Hint: Find a subset $\ A \subset \mathbb{N}\ $ with the property that, for each $\ n \in \mathbb{N},\ $ the set $ \{kn\}_{k \in \mathbb{N} }$ has finitely many elements in common with $A.$

Answer: The prime numbers is the standard example of such a subset of $\mathbb{N}$ with the property in the hint. This point is to make the 2nd, 3rd, 5th, 7th, ... members of our sequence have one sign, and every other member of our sequence have another sign. For example, $$ a_1 = \frac{1}{1},\ a_2 = \frac{-1}{4},\ a_3 = \frac{-1}{4},\ a_4 = \frac{1}{3},\ a_5 = \frac{-1}{4},\ a_6 = \frac{1}{5},\ a_7 = \frac{-1}{6},\ a_8 = a_9 = a_{10} = \frac{1}{3} \times \frac{1}{7},\ a_{11} = \frac{-1}{8},\ ... $$ This is simply the alternating harmonic series split up in a way that (I thought) answers the question.

That was my answer, but now I'm not sure the series above is a valid counter-example. I constructed it so that every subseries of the form $\displaystyle\sum _{n=1}^\infty a_{kn}$ contains finitely many negative numbers. And then I'm not sure what my logic was. It was either that "every subseries of a monotonic divergent series diverges", but now I realise that this is not what the link says, and is also not true. Or my reasoning was that, for each $k \in \mathbb{N},\ $ every subseries $\displaystyle\sum _{n=1}^\infty \frac{1}{kn}$ of the harmonic series, diverges, which is true, but the subseries of my series defined above do not match these subseries of the harmonic series.

So the problem remains open...

  • $\begingroup$ Do you mean the sub-series converges absolutely? Otherwise, the question does not make sense. For, $\sum_{k=2}^\infty a_k$ is a convergent sub-series of the original series. $\endgroup$ Feb 7, 2021 at 4:35
  • $\begingroup$ @Danny I suspect you misread $\displaystyle\sum_{n=1}^\infty a_{kn}\ $ as $\displaystyle\sum_{n=1}^\infty a_{n_k}\ $. Am I right? (If so, it’s an understandable mistake as they do look similar). The first series (the ones of particular interest in the question) is something like $\displaystyle\sum_{n=1}^\infty a_{8n}.$ $\endgroup$ Feb 7, 2021 at 8:57

2 Answers 2


Your primes idea is a good one! It just takes a little bit of finagling to make it clearer that the idea works. In particular, it relies on some nontrivial number theory -- but this kind of interplay is part of the fun of mathematics!

Here's the plan: We'll use the harmonic series on the composite numbers. This normally diverges (of course) but since there are infinitely many primes, we have infinitely many chances to knock it back down to $0$. Then we'll show that this converges (it's obvious that it does not converge absolutely). Lastly, we'll use your idea about arithmetic progressions only intersecting the primes finitely many times to conclude.

To be clear, our sequence will be:

$$ \begin{array}{|c|c|} \hline \mathbf{n} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \mathbf{a_n} & 1 & -1 & 0 & \frac{1}{4} & -\frac{1}{4} & \frac{1}{6} & -\frac{1}{6} & \frac{1}{8} & \frac{1}{9} & \frac{1}{10} & -(\frac{1}{8} + \frac{1}{9} + \frac{1}{10}) & \frac{1}{12}\\ \hline \mathbf{\sum_{i \leq n} a_i} & 1 & 0 & 0 & \frac{1}{4} & 0 & \frac{1}{6} & 0 & \frac{1}{8} & \frac{1}{8} + \frac{1}{9} & \frac{1}{8} + \frac{1}{9} + \frac{1}{10} & 0 & \frac{1}{12} \\ \hline \end{array} $$

and so on.

It's obvious that the sum tends to stay around $0$, but it's not obvious that it converges to $0$. After all, who's to say we won't have a long dry spell between primes where we move above $10^{-100}$? This doesn't need to happen often. It only needs to happen infinitely many times, and convergence is ruined.

Thankfully, this can't happen. It's a corollary of the prime number theorem that for any $\epsilon$ we like, the distance between primes is eventually $\epsilon$ proportionally small. You can read more about this on wikipedia, but I've transcribed the relevant fact here:

For all $\epsilon > 0$, for all large $k$ (where "large" depends on $\epsilon$), we have $p_{k+1} - p_k \leq \epsilon p_k$

Why does this guarantee convergence? Well, let $\epsilon > 0$. We want to show that, eventually, we stay within $\epsilon$ of $0$. But if $n$ is bigger than $p_k$ (as guaranteed by the above theorem), we see:

$$ \begin{aligned} \sum_{i \leq n} a_i &\leq \sum_{i < p_{l+1}} a_i \\ &\leq H(p_{l+1}) - H(p_l) \\ &\sim \log(p_{l+1}) - \log(p_l) \\ &= \log(\frac{p_{l+1}}{p_l}) \\ &\leq \log(\frac{p_l + p_l \epsilon}{p_l}) \\ &= \log(1 + \epsilon) \\ &\leq \epsilon \end{aligned} $$

Here $p_l$ is the prime immediately before $n$, and $H(m)$ is the $m$th harmonic number.

But then for $n$ large, $\sum_{i \leq n} a_i \leq \epsilon$, and our series converges (conditionally) to $0$.

Now we can finish up! As you noticed, any sequence $(kn)_{n \in \mathbb{N}}$ can only intersect the primes finitely many times (indeed, at most once). So for any sequence of that form, we see $\sum_n a_{kn}$ contains a tail of the harmonic series (I guess we're off by a factor of $\frac{1}{k}$, but what's a constant between friends?), and diverges.

Thanks for asking this -- it was a lot of fun!

I hope this helps ^_^


(This is a simplified version of my previous solution when I used harmonic series to fill in "gaps") The statement is not true.
To see this fix a increasing sequence of primes $p_{l}$ such that $lim_{l\to\infty} (p_{l+1} - p_{l}) = \infty$ and define $a_{n}$ such that $a_{n} = \frac{1}{l}$ if $n=p_{l}$.
To fill in the gaps in the definition of $a_{n}$ where n is between $p_{l}$ and $p_{l+1}$ set $a_{n} = - \frac{1}{l*(p_{l+1} - p_{l})}$ In words, $a_{n}$ is positive only for $n=p_{l}$ and its value for such $n$ is $\frac{1}{l}$; then it is followed by small, negative, and identical entries which add up to $\frac{1}{l}$ and then again $a_{n}$ for $n= p_{l+1}$ which is $\frac{1}{l+1}$ followed by negative entries adding up to $\frac{1}{l+1}$. We note that the sum of terms in "negative block" is $\frac{1}{l}$ hence sum of a sub-block which is represented by an arithmetic progression with difference $k$ which fall into such block is $\frac{1}{lk}$.
Lastly, if we take a sub series of the form $a_{nk}$ then indexes are never primes so all terms are negative. Further, this sub series, on each sub block adds up to $-\frac{1}{kl}$ and when summed up by l is divergent.
Addendum, please note that the original series $\sum a_{n}$ is convergent. This follows easy by looking at the difference between partial sums (i.e. checking that partial sums forms Cauchy sequence).

  • $\begingroup$ So your first step is to let $a_2 = \frac{1}{1}, a_3 = \frac{1}{2}, a_5 = \frac{1}{3}, a_7 = \frac{1}{4}, a_{11} = \frac{1}{5}, ... .\ $ Your next step is to "take as many terms from the remaining sequence of $\frac{1}{m}\ $ as possible. I don't get this: there aren't any terms from $\frac{1}{m}\ $ remaining: you've used them all! $\endgroup$ Feb 7, 2021 at 10:45
  • $\begingroup$ @Adam no, you build the sequence by induction sort of. There is never "nothing left" at any stage. Also, there is nothing wrong when some terms of the series are are repeated. Will try to improve wording anyhow. Thanks for the comment. $\endgroup$
    – Salcio
    Feb 7, 2021 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy