# Are there any series whose convergence is unknown?

Are there any infinite series about which we don't know whether it converges or not? Or are the convergence tests exhaustive, so that in the hands of a competent mathematician any series will eventually be shown to converge or diverge?

EDIT: People were kind enough to point out that without imposing restrictions on the terms it's trivial to find such "open problem" sequences. So, to clarify, what I had in mind were sequences whose terms are composed of "simple" functions, the kind you would find in an introductory calculus text: exponential, factorial, etc.

• Let $a_n = 1$ if $p_n+2$ is a prime where $p_n$ is the $n$th prime. The convergence of $\sum_n a_n$ is equivalent to the twin prime conjecture. Of course, this isn't the likely sense in which the OP asked the question but maybe the question needs to be more specific? Feb 5, 2011 at 17:36
• The problem with this question is that one can encode a great deal of other mathematics as the question of whether some artificial series converges, as Dinesh as shown. One has to somehow restrict to a natural series, and it's far from clear what this even means. Feb 5, 2011 at 17:44
• @Dinesh: In fact any statement on proving something is countably infinite could be restated using an indicator function like what you have stated. Anyway good one.
– user17762
Feb 5, 2011 at 17:47
• Thanks guys, all good points. I clarified things a bit. (My question reminds me of the Ali G interview where he asks "Will a computer ever be able to work out what is 999999999... multiplied by 999999...?".) Feb 5, 2011 at 21:55
• The answer to whether a computer can work out... is you are asking for $(10^m-1)*(10^n-1)=10^{(m+n)}-10^m-10^n+1$ and I think Wolfram Alpha can do that already. Jul 2, 2011 at 3:40

It is unknown whether $$\sum_{n=1}^\infty\frac{1}{n^3\sin^2n}$$ converges or not. The difficulty here is that convergence depends on the term $$n\sin n$$ not being too small, which in turn depends on how well $$\pi$$ can be approximated by rational numbers. It is possible that, if $$\pi$$ can be approximated `too well' by rationals, then this will diverge. See this MathOverflow question for a discussion of this particular series.

Another even simpler example of a sequence (no summation) for which it is not known whether it converges or not is $$x_n=\frac{1}{n^2\sin n}.$$ We would expect this to tend to zero, but the proof is beyond what is currently known. Suppose that there were only finitely many rational numbers $$p/q$$ with $$\vert p/q-\pi\vert\le q^{-3+\epsilon}$$ (for any $$\epsilon > 0$$), then $$x_n$$ would tend to zero at rate $$O(n^{-\epsilon})$$. If, on the other hand, there were infinitely many rationals satisfying $$\vert p/q-\pi\vert\le q^{-3-\epsilon}$$, then infinitely many $$x_n$$ would be of order at least $$n^\epsilon$$, so it diverges. This can be expressed in terms of the irrationality measure of $$\pi$$. The sequence $$x_n$$ converges to zero if the irrationality measure of $$\pi$$ is less than 3, and diverges if it is geater than 3. Currently, the best known bound for the irrationality measure is that it is no more than about $$7.6063$$* (see the link to the mathworld page above). It is expected that the irrationality measure of $$\pi$$ is 2 (it is known that all but a zero-measure set of real numbers have irrationality measure 2). Therefore, it is expected that $$x_n$$ tends to zero, but there is currently no proof of this.

[* The best known bound for the irrationality measure, as listed on the mathworld page, has been improved! It is now 7.10320533, according to the (not yet peer-reviewed) paper by Zeilberger and Zudlin, 2019. This is still much too small an improvement to say whether or not the sequence above converges.]

• Mentioned at MO mathoverflow.net/questions/100265/… Jun 22, 2012 at 13:16
• I find it quite incredible that such a simple sequence ($x_n$) has unknown convergence.
– Anon
Jan 6, 2017 at 2:32
• Also mentioned on Wikipedia and named “Flint Hills series” both there and in the MO post you referenced.
– MvG
Jan 6, 2017 at 8:47
• I am blown away by this. Nov 1, 2017 at 15:04
• @GeorgeLowther I know it is an old post, but could you briefly explain how to obtain that $x_k=O(n^{-\varepsilon}$ in the case when there are only finite many rationals satisfying the inequality? I see how the existance of infinitely many of them implies the divergence by bounding from below a subsequence of $x_n$ that diverges, but I cannot bound the sequence above to prove the convergence in the other case. Jun 24, 2020 at 7:24

It is unknown whether the series: $$\sum_n \frac{(-1)^n n}{p_n}$$ converges. Here, $p_n$ is the $n$-th prime number. This problem is posed in Guy's book on unsolved problems in number theory and I am pretty sure that it originated from Erdős.

• Very nice. +1 for Erdős. Feb 19, 2016 at 10:52
• Huh, well, if you accept the prime number theorem, then this converges, else, it is unknowable. Mar 9, 2017 at 14:48
• @SimplyBeautifulArt the prime number theorem does not imply convergence of this series. Mar 13, 2017 at 2:38
• Really? By the alternating test, all I need to do is show that $\frac n{p_n}\to0$. Mar 13, 2017 at 12:36
• @SimplyBeautifulArt in order to use the alternating series test you must show that the terms of the series are monotonically decreasing. Mar 17, 2017 at 21:30

As kind of a joke answer, but technically correct, and motivated by Chandru's deleted reply, $$\sum_{n=0}^\infty \sin(2\pi n!\,x)$$ where $x$ is the Euler-Mascheroni constant, or almost any other number whose rationality has not been settled. (If $x$ is rational, the series converges. The implication does not go the other way.)

• Out of curiosity, what makes this a "joke" answer? May 17, 2013 at 1:58
• There is no analysis of decay rate involved. May 17, 2013 at 19:20
• The comma before "or almost any" confused me for a second, better to remove it imho, without it one understands immediately your use of x instead of the usual gamma. (there is an Oxford comma, but not really for "or") Feb 26, 2018 at 6:50

The Riemann hypothesis is that $$\sum_{n=2}^\infty \frac{\Lambda(n)-1}{n^{1/2}\log^{3+\epsilon} n}$$ converges for any $$\epsilon > 0$$ (see here for a discussion of that $$\epsilon$$).

• Here, $\Lambda(n)$ is the logarithm of the prime number $p$ such that $n$ is a power of $p$, if any, or $0$ if $n$ is not a prime power. (It says so in the first link, but I thought it helpful to have it on this page.) I also presume that all of the logarithms are natural logarithms. Feb 9, 2018 at 21:54
• It is standard in analytic number theory to have all logs implicitly natural. Feb 26, 2018 at 16:39

Let

$$h(x) = \sum_{k=1}^\infty \frac{\{2^kx\}-\frac{1}{2}}{k},$$

where the curly brackets represent the fractional part function, and $$x\in [0,1]$$. For almost all numbers $$x$$, it is not know whether the series converge or not. It is believed to converge for almost every number, yet it is very hard to come up with a single non-rational number, not one that is artificially manufactured, such that the convergence status is known. If for instance you can prove that it converges for $$x = \frac{\pi}{4}$$ (everyone strongly believes that it does, and it is backed by empirical evidence), you would instantly become very famous in the math community. Chances are that it is impossible to prove or disprove convergence for $$x=\pi,e,\log 2$$ and most other math constants. This question was raised in section 4.3.(a) in this article, where you can find more details about it.

It would be interesting to study the wildly erratic behavior of this function, which is not only discontinuous everywhere, but admits a dense set of singularities (where it does not converge.)

• it seems the link to the article has rotted... do you still remember the title?
– D.R.
Feb 20 at 19:28
• @D.R. Sorry about that. I just got it resuscitated: vgranville.com/2022/02/… Feb 20 at 20:34

From Wikipedia:

The statement that the equation:

$$\frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}$$

is valid for every $$s$$ with real part greater than 1/2, with the sum on the right hand side converging, is equivalent to the Riemann hypothesis.

This in turn can be rewritten as:

$$\displaystyle \frac{1}{\lim\limits_{k\to \infty } \, \left(\sum\limits_{a=1}^{k} \frac{1}{a^s}+\frac{1}{(s-1) k^{s-1}}\right)} =$$ $$\displaystyle \lim_{k \rightarrow \infty} \left( \underbrace{1 - \sum_{2 \leq a \leq k} \frac{1}{a^{s}} + \underset{ab \leq k}{\sum_{a \geq 2} \sum_{b \geq 2}} \frac{1}{(ab)^{s}} - \underset{abc \leq k}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2}} \frac{1}{(abc)^{s}} + \underset{abcd \leq k}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2} \sum_{d \geq 2}} \frac{1}{(abcd)^{s}} - \cdots}_{\text{number of alternating sums} > \frac{\log(k)}{\log(2)}} \right)$$

where:

$$\Re(s)>0:\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\zeta(s)=\lim\limits_{k\to \infty } \, \left(\sum\limits_{n=1}^{k} \frac{1}{n^s}+\frac{1}{(s-1) k^{s-1}}\right)$$

Reuns answer would be: $$\displaystyle \sum_{n=2}^{\infty} \left( \underbrace{-\frac{1}{\sqrt{n} \log^{3+\epsilon}(n)}+\underset{a = n}{\sum_{a \geq 2}} \frac{\log(a)}{\sqrt{n}\log^{3+\epsilon}(n)} - \underset{ab = n}{\sum_{a \geq 2} \sum_{b \geq 2}} \frac{\log(a)}{\sqrt{n}\log^{3+\epsilon}(n)} + \underset{abc = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2}} \frac{\log(a)}{\sqrt{n}\log^{3+\epsilon}(n)} - \underset{abcd = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2} \sum_{d \geq 2}} \frac{\log(a)}{\sqrt{n}\log^{3+\epsilon}(n)} + \cdots}_{\text{number of alternating sums} > \frac{\log(n)}{\log(2)}} \right)$$

Mathematica: https://pastebin.com/gxAE6ZgY

• Jun 14, 2021 at 8:47

The Riemann hypothesis is equivalent to: $$\lim_{n\to \infty } \, \frac{\sum\limits_{k=1}^n \lambda (k)}{n^{\frac{1}{2}+\epsilon}}=0$$ where $$\lambda(k)$$ is the Liouville Lambda function.

This in turn can be rewritten as:

$$\lim\limits_{n \rightarrow \infty}\frac{1}{n^{\frac{1}{2}+\epsilon}}\left(\underbrace{\underset {a^2 \leq n} {\sum_ {a\geq 1}} 1 - \underset {a^2 b \leq n} {\sum_ {a\geq 1}\sum_{b\geq 2}} 1 + \underset {a^2 bc \leq n} {\sum_ {a\geq 1}\sum_ {b\geq 2}\sum_ {c\geq 2}} 1 - \underset {a^2 bcd \leq n} {\sum_ {a\geq 1}\sum_ {b\geq 2}\sum_ {c\geq 2}\sum_ {d\geq 2}} 1 + \cdots}_{\text{number of alternating sums} > \frac{\log(n)}{\log(2)}}\right)=0$$