149
$\begingroup$

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.

  • 1
    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.2011-07-02

6 Answers 6

189

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.]

  • 1
    I am blown away by this.2017-11-01
59

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.)

  • 0
    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")2018-02-26
56

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.

  • 14
    @SimplyBeautifulArt in order to use the alternating series test you must show that the terms of the series are monotonically decreasing.2017-03-17
21

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 $\epsilon$).

  • 5
    It is standard in analytic number theory to have all logs implicitly natural.2018-02-26