# Series whose convergence is not known

For most of the mathematical concepts I learn, it has more or less always been possible to find (at least google and find) unsolved problems pertaining to that specific concept. Keeping a bag of unsolved problems on most topics I know has been to my benefit in that it reaffirms me that mathematics is a thriving subject.

Coming to point, I am unable to find an elementary series of the kind we know on real analysis courses whose convergence is an unsolved problem. Please, share if you have any.

Thanks.

• Here's an example mathoverflow.net/questions/24579/convergence-of-a-series Commented May 24, 2011 at 14:58
• @Andrey,thank you. Sigh, that means my habit is worth advancing. Commented May 24, 2011 at 15:23
• I think it should be quite easy to write many such series. Consider e.g. $\sum_{k=1}^\infty (-1)^{p_k}/k$, where $p_k$ is the $k$-th digit of $\pi$ (or the digits of $\pi^e+\gamma^\pi$, just in case) Commented May 24, 2011 at 16:06
• There was a very similar question on math.SE a while ago, which I posted an answer to. math.stackexchange.com/q/20555/1321 Commented May 24, 2011 at 21:39
• For example, does $(n^2\sin n)^{-1}\to0$ (very likely, but it's unknown). Similarly does $\sum_n(n^3\sin n)^{-1}$ converge? Again, it probably does, but is impossible to say for sure without better bounds on the irrationality measure of $\pi$ than are currently available. Commented May 24, 2011 at 21:44

$1/\zeta(s)=\sum_{n>0}\frac{\mu(n)}{n^s}$ where $\mu$ is the Moebius function. This series is known to converge for $s\ge 1$ and diverge for $s\le 1/2$. Its convergence is unknown if $1/2< s< 1$ (convergence in this interval is essentially the Riemann hypothesis).

• In addition, convergence at $s = 1$ is equivalent to the prime number theorem. Commented May 25, 2011 at 10:01
• It is very exciting to see such engaging list of answers have been waiting me as the Internet connection failed terribly for about a day. Commented May 26, 2011 at 16:08
• @Richard Borcherds, thank you very much. Commented May 26, 2011 at 16:09
• @FranzLemmermeyer: Does the other known endpoint $s=1/2$ also correspond to some theorem? Commented Oct 6, 2013 at 9:41
• @VítTuček Don't quote me on that, but I believe we can infer from this that the error term in PNT is not $O(n^{1/2-\varepsilon})$ for any $\varepsilon>0$. Commented Mar 10, 2016 at 21:01

For an interesting example, take $\sum_{n=1}^\infty \frac{|\sin(n)|^n}{n}$. Deciding whether or not this converges seems to require more knowledge than is currently available about the rational approximations of $\pi$. The series $\sum_{n=1}^\infty \frac{|\sin(n t \pi)|^n}{n}$ converges for almost every real $t$ (in the sense of Lebesgue measure), but diverges for $t$ in a dense $G_\delta$ subset of $\mathbb R$.

EDIT: ... and now it is known to converge, as Sam Hopkins commented.

• If $t = p/(2q)$ where $p$ and $q$ are odd integers, then $|\sin(nt\pi)|=1$ whenever $n$ is an odd multiple of $q$, and so the series diverges for such $t$. Such $t$ form a dense subset of $\mathbb R$. The set where a series of nonnegative continuous functions diverges is a $G_\delta$, so we have a dense $G_\delta$. For the convergence a.e., note that $$\int_0^1 \frac{|\sin(nt\pi)|^n}{n}\ dt \sim \frac{C}{n^{3/2}}$$ so the series converges in $L^1[0,1]$, therefore the sum is finite almost everywhere in $[0,1]$ (and, by periodicity, in $\mathbb R$). Commented Mar 11, 2012 at 18:13
• Sorry for bumping, but do you have a reference for the first series? There's a bounty on it here math.stackexchange.com/questions/823816/…
– user23855
Commented Jun 16, 2014 at 14:55
• I'm not convinced by what's at that link. Since $\sum_n |\sin(nt\pi)|^n/n$ diverges for $t$ in a dense $G_\delta$, and in particular for uncountably many irrational $t$, it seems to me you need to use more than just Weyl equidistribution. Commented Oct 19, 2016 at 7:31
• If you're just saying that the series appears to converge, then I agree: I would be very surprised if it did not converge. But that's not a proof. Commented Oct 19, 2016 at 18:19
• Commented Sep 29, 2017 at 18:33

First, here's a silly example: define $a_n$ to be 1 if $n$ and $n+2$ are both prime. I think it's fair to say that that doesn't count as the kind of series that crops up in an elementary analysis course. But the real reason it's silly is that it is just an encoding of a problem that isn't about the convergence of a series at all.

Somehow I feel that the examples involving $\sin n$ are of a similar flavour, even though they resemble elementary series in analysis much more closely. When you try to solve them, you quickly find that the problem turns into something else (in this case, questions about rational approximations to $\pi$).

It's probably asking for too much, but I wonder whether there is a series whose convergence is not known, where one wouldn't get the feeling that the convergence of the series was just an artificial way of asking a different problem. The reason it may be asking too much is that any attempt to prove convergence is likely to involve a certain amount of reformulation, and who is to say what counts as changing the problem to a different one?

But let me attempt to ask a narrower question that gets somewhere in the right direction. Is there a series $(a_n)$ that has a relatively simple analytic definition (so you can't just encode some counting problem), such that every $a_n$ is non-negative, the $a_n$ are decreasing, the difference sequence $a_{n+1}-a_n$ is decreasing, the difference sequence of that is decreasing, and so on, and the convergence of the sum $\sum a_n$ is unknown?

• @Gowers, thank you for your re-framing of the question. (for the $a-n$ series, it is trivially to be completed with $a_n$ 0 when at least one is not prime... ) I was thinking of series whose terms are expressible in terms of elementary functions and as you said it rightly, not encoding a counting problem. Commented May 24, 2011 at 18:14
• I feel like one would need to define the word "unknown" too: there are many pairs of integer numbers whose prime factorization is unknown (in the sense that no human being has ever done it and, in all likelihood, no one will ever be able to do it in the foreseeable future). Still, we have a simple finite algorithm and we are (sort of) happy. On the other hand, I haven't seen a computer program that can figure the convergence question out given the symbolic definition that goes beyond the Hardy domain (say, recursive) and I can create a recurrent formula that nobody has ever looked at. Commented May 24, 2011 at 18:15
• I meant the $a_n$ series. Commented May 24, 2011 at 18:19
• To clarify what I said: let $f$ be a nice decreasing "explicit" function tending to a limit (even in Hardy domain) for which it is hard to determine whether it is eventually negative. Put $a_n=\max(0,f(n))$ and enjoy. As far as I know, you can do elementary functions but not antiderivatives. Take, say, $f(t)=-0.4+\int_{-\infty}^t e^{-x^4-3x^2-1}dx$ and go figure (if you have too much confidence in your CAS, I'll write something in the same spirit but more complicated, subtracting another integral instead of a constant. Commented May 24, 2011 at 19:40
• But this example is too easy: Maple's numerical integration says $\int_{-\infty}^\infty e^{-x^4-3x^2-1} dx = .3538037606$, and it would be straightforward if tedious to use estimates to prove rigorously that this integral is less than 0.4. Oh, and Maple can also get a closed form for the integral: it is $\frac{\sqrt{3}}{2} e^{1/8} K(1/4,9/8)$ where $K$ is a Bessel function of the second kind. Commented May 25, 2011 at 17:39

Convergence of $\sum_{k=1}^\infty (-1)^k \frac{k}{p_k}$ is unknown, where $p_k$ is k-th prime (Guy 1994, p. 203; Erdős 1998; Finch 2003, according to Eric Weisstein's MathWorld).

Let $B_{k}$ refer to the Bernoulli numbers. Then consider the series

$$\sum_{n=0}^{\infty}\left(n+\frac{1}{2}\right)\biggr|\sum_{k=0}^{\infty}\frac{c_{2n+1,2k+1}}{2k+2}\log\left(\frac{2k+1}{2k+2}\frac{(-1)^{k}B_{2k+2}(2\pi)^{2k+2}}{2(2k+2)!}\right)\biggr|^{2}$$

where

$$c_{2n+1,2k+1}=\frac{(-1)^{n-k}(2n+2k+2)!}{2^{2n+1}(n-k)!(n+k+1)!(2k+1)!}.$$

It is not known whether this series converges. In fact, its convergence is equivalent to the Riemann Hypothesis.

(There are a lot of ways to make series that converge if and only if RH)

Hope that helps,

Reference:

(1) John Corning Carey's Paper: http://jcarey.best.vwh.net/RHHardy.pdf

• Oh thanks. Googling with you term "onvergence is equivalent to the Riemann Hypothesis" helps too. (search gave this www.math.lsa.umich.edu/~lagarias/doc/mt-holyoke-rev.pdf) Commented May 24, 2011 at 15:21
• @Chulumba: Another that comes to mind depending on whether you call $\psi(x)$ elementary is the series $$\sum_{n=1}^{\infty}\frac{|\psi(n)-n|}{n^{\frac{3}{2}+\epsilon}}.$$ We do not know its convergence for any $0<\epsilon<1/2$. Commented May 24, 2011 at 15:26
• @Eric, thanks again for the reference. Commented May 24, 2011 at 15:29

You probably didn't intend this, but there are plenty of artificial ways to write down series whose convergence is equivalent to various types of unsolved problems. For example, let $S$ be any set such that it is an unsolved problem whether $S$ is infinite or not, and consider the convergence of the series $\sum_{s \in S} 1$.

• @Qiaochu Yuan, I would be happy to hear what (if any) unsolved problem that series encoded. Commented May 24, 2011 at 18:14
• So it is the same(generalized) as Gower's above. Commented May 24, 2011 at 18:21
• @Chulumba: for example (again, this is really artificial) let $S$ be the set of non-trivial zeroes of the zeta function not on the critical line. Commented May 24, 2011 at 19:27
• Consider $\sum a_n x^n$ where $a_n$ counts the number of planar graphs on $n$ vertices which are not $4$-colorable, or the proofs that $P\ne NP$ of length $n$, or the number of $\sqrt{n} \times \sqrt{n}$ matrices whose entries are $0$ or $1$ with no $1$s which are horizontally or vertically adjacent. The convergence of the first is known but hard, the second is open, and the third is known to converge for some $x$, to diverge for some $x$, and it's open for some $x$, since the hard square constant is not known exactly. Commented May 25, 2011 at 5:27

Assume that $$A\subset\mathbb{N}$$ contains no 3-term arithmetic progression. It was conjectured by Erdős and Turán that $$\sum_{a\in A}\frac{1}{a}$$ converges. As far as I know this is open, although we are getting closer to a proof, see Sanders' paper.

• I think to stay in the spirit of the original question, you'd have to specify a particular set $A$ containing no 3-term arithmetic progression for which it's not clear whether the sum converges. Do you have any examples of that? For example, what about the greedy sequence (A003278 in the OEIS)? Commented May 25, 2011 at 20:22
• @Robert: Your criticism is fair, and I don't know about a specific $A$. I think it is very difficult to construct large $A$'s with no 3-term arithmetic progression, the best examples go back to Behrend and for them $\sum_{a\in A}\frac{1}{a}$ is known to converge. Also, clearly, the OP's question is vague: what does it mean to "find" or "give" a sequence, what is "elementary" etc. Commented May 25, 2011 at 23:08
• @Robert: The Erdős-Turán conjecture is equivalent to convergence of $\sum_{i=1}^{\infty} \frac{r_k([1,2^n])}{2^n}$, where $r_k(A)$ is the size of largest subset of $A$ without an arithmetic progression of length $k$. Source: exercise 10.0.6 in Additive Combinations by Tao and Vu. Commented Jul 2, 2013 at 13:37

This is a problem about convergence of a certain type of measures, I am somewhat familiar with, which is yet to be proved:

Let $$T = \sum_{j=1}^k Q_j \frac{d^j}{dx}$$ with $\deg Q_j\leq j$ with equality for at least one $j,$ but $\deg Q_k < k.$ There are unique monic eigenpolynomials $p_n$ of each degree $n>n_0,$ that is $Tp_n = \lambda_n p_n.$

Let $d := max_{j \in [j_0,k]} \frac{j-j_0}{j - \deg Q_j}$ where $j_0$ is the largest $j$ such that $\deg Q_j = j,$ and consider the polynomials $q_n(z)=p_n(n^dz).$

Conjecture: the root measures $\mu_n$ obtained by placing a point mas of weight $1/n$ at each root of $q_n$ converges weakly to a probability measure $\mu$ with has a compact support in the shape of a tree.