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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
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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
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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
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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

3 Answers 3

up vote 60 down vote accepted

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.)

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Out of curiosity, what makes this a "joke" answer? –  AlanH May 17 '13 at 1:58
    
There is no analysis of decay rate involved. –  André Nicolas May 17 '13 at 19:20

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.

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