# 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.

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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? –  Dinesh Feb 5 '11 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. –  Qiaochu Yuan Feb 5 '11 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 '11 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...?".) –  whitman Feb 5 '11 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. –  Ross Millikan Jul 2 '11 at 3:40
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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' enough 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.

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Mentioned at MO mathoverflow.net/questions/100265/… –  Alexander Chervov Jun 22 '12 at 13:16
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.)
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 Erdos.