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If you have a series of numbers (where the first number corresponds to the index zero):

$$ \frac{1}{2}+ \frac{1}{24}+ \frac{1}{720}+... $$

Is there a systematic way to work backwards and find a recurrence relation? Or in other words, something that would be easier for a computer to understand (let's say I wanted to write a program which can take a series of numbers and then compute the relation automatically) $a_n=...$

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No, since there are infinitely many infinite series that have the same first three terms. It’s really a problem in pattern recognition: you’re being asked to spot the pattern that the person who posed the question had in mind, which will typically be some fairly simple, regular pattern. Here the expectation is that you will recognize the denominators as factorials, specifically, as $2!,4!$, and $6!$. If you do, it becomes clear that the $k$-th term is $\frac1{(2k+2)!}$. If $a_k$ is the $k$-th term, then

$$a_k=\frac{a_{k-1}}{(2k+1)(2k+2)}\;.$$

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$2!$ $4!$ $6!$ Hmm, do I see a pattern?

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  • $\begingroup$ You've nailed it. I did not see this pattern. I wonder if there is another method besides 'recognizing'? Maybe something that would be more friendly for a computer to understand. $\endgroup$ – Bob Shannon Dec 8 '13 at 4:20
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    $\begingroup$ If there were a positive answer to your question, many of us would be out of a job :( $\endgroup$ – Igor Rivin Dec 8 '13 at 4:22
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Don't believe what they tell you, the real answer is $$ a_n=\frac{1}{337n^2-989n+654} $$ ;-)

P.S. Actually, I can tell you how to write a computer program which will solve all these questions without having to guess anything. This is called Lagrange interpolation. But I doubt that it is the expected answer.

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  • $\begingroup$ Wow, look at that, it holds for all $n \ge 1 $...nice. Will have to read up on that. Thank you. $\endgroup$ – Bob Shannon Dec 8 '13 at 4:42
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If course the only valid answer is (2*p)! in the denominators, where p runs through the non-composite positive integers, p=1,2,3,5,7,11,13,... :-)

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