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Is there any simple series where all the usual convergence tests are inconclusive? And how is convergence/divergence determined for these series?

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Sure, but this is a bit like asking "what's the largest number anyone has ever used?" If you use that number, you can always increment it. Create a series all the known tests fail for and someone will create a new test to deal with that series. –  Ryan Budney Jun 24 '12 at 20:46
    
@RyanBudney Unless of course, creating a test to deal with that series is extremely difficult. I think this is a good question. –  Ragib Zaman Jun 24 '12 at 20:47
    
By usual, I mean en.wikipedia.org/wiki/Convergence_tests –  user1708 Jun 24 '12 at 20:48
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It would be more interesting to know if there is some positive monotonically decreasing series. And where its not an open problem. –  user1708 Jun 24 '12 at 21:13

6 Answers 6

In a spirit similar to Henning's answer: let $p_n$ denote the $n$th prime number, and set $a_n = 1$ if $p_n$ and $p_n + 2$ are both prime, and $0$ otherwise. Divergence of the series is of course equivalent to infinitude of twin primes. In fact, any statement about infinitude of subsets of countable sets can be mapped to an equivalent statement about convergence or divergence of infinite series.

Now, after massaging the statement above and recalling Godel's incompleteness theorem, one could see that in fact there are infinitely many infinite series whose convergence or divergence could not be determined given our (finite) axiomatic system.

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How about $$a_n = \begin{cases} 0 & \text{if $2n$ is the sum of two primes}\\ 1 & \text{otherwise}\end{cases} $$ A nontrivial amount of fame awaits you if you can determine whether that converges or not.

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It diverges. :D –  user31373 Jun 24 '12 at 20:49
    
I think you mean $\sum_{n=1}^\infty (1-a_n)$. –  Robert Israel Jun 24 '12 at 20:50
    
@LeonidKovalev: Happier now? –  Henning Makholm Jun 24 '12 at 20:52
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@HenningMakholm Exactly the opposite. I thought I was going to be famous. :-( –  user31373 Jun 24 '12 at 20:54
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Even if such a series converges, all we conclude is that there is at most a finite number of counterexamples to the Goldbach conjecture. Another example to convey the idea is $a_n= 1$ if $p_n+2$ is prime, 0 otherwise. –  Ragib Zaman Jun 24 '12 at 20:54

There are a lot of series for which we do not even know whether they converge or not. For instance, the convergence of the series $$\sum_{n} \dfrac{\mu(n)}{n^s}$$ for $\text{Re}(s) \in (1/2,1)$ is essentially equivalent to the Riemann Hypothesis. Hence, new techniques and ideas are needed to analyze convergence in this case.

There are plenty of other examples as well where we do not know about the convergence or divergence of the series. A trivial example is as follows. Any conjecture on infinitude of a certain thing can be converted into a series. For instance, if there is a conjecture stating there are infinitely many $x \in X$ such that a certain statement $P(x)$ is true. Then this can be converted into a series $$\sum_{x \in X} \mathbb{1}_{P(x)\text{ is true}}$$ and analyzing the convergence/divergence of this is equivalent to the conjecture being true or false.

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Related math.stackexchange.com/questions/20555/… –  user17762 Jun 24 '12 at 20:59

For series with positive terms, Kummer's test always determines convergence. You will find this less exciting after seeing what the test is.

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It depends on what you think of as a "simple" series. There are many series where the convergence/divergence is unknown and extremely difficult. Here is a very relevant link. There are examples of series whose convergence is equivalent to the Riemann hypothesis and other famous conjectures.

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Here is a problem that i recently had at university (for homework). The problem is to determine the convergence of $\sum_{n \ge 1} \frac{\cos(\sqrt{n})}{n}$Here is an example

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