Need help with a sum series (Frobenius series)

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SUMMARY

The discussion focuses on deriving a condensed form of a Frobenius series solution for a recurrence relation defined as a_{n+2} = \frac{(n-1)(n-2)-\frac{2k}{w_{o}^{2}}}{R^{2}(n+1)(n+2)}a_{n}. The user successfully expanded the series to f(r) = 1 + \frac{-\frac{2k}{w_{o}^{2}}}{6R^{2}}r^{2} + \frac{-\frac{20k}{w_{o}^{2}}+\frac{4k^{2}}{w_{o}^{4}}}{120R^{4}}r^{4} + \frac{-\frac{560k}{w_{o}^{2}}+\frac{152k^{2}}{w_{o}^{4}}-\frac{8k^{3}}{w_{o}^{6}}}{5040R^{6}}r^{6} and identified the factorial pattern in the denominators. The user seeks assistance in condensing the numerator into a single series term, proposing a potential solution involving the double factorial and a summation.

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Students and researchers in mathematics, particularly those focusing on series solutions and recurrence relations, as well as anyone working on problems involving Frobenius series expansions.

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Homework Statement


I've obtained a recurrence relation of:
[itex]a_{n+2}[/itex] = [itex]\frac{(n-1)(n-2)-\frac{2k}{w_{o}^{2}}}{R^{2}(n+1)(n+2)}[/itex][itex]a_{n}[/itex]
from a Frobenius series solution problem and I've expanded it to give the series:

f(r) = 1 + [itex]\frac{-\frac{2k}{w_{o}^{2}}}{6R^{2}}[/itex][itex]r^{2}[/itex] + [itex]\frac{-\frac{20k}{w_{o}^{2}}+\frac{4k^{2}}{w_{o}^{4}}}{120R^{4}}[/itex][itex]r^{4}[/itex] + [itex]\frac{-\frac{560k}{w_{o}^{2}}+\frac{152k^{2}}{w_{o}^{4}}-\frac{8k^{3}}{w_{o}^{6}}}{5040R^{6}}[/itex][itex]r^{6}[/itex] ...

I can see that the denominator is 3![itex]R^{2}[/itex] , 5![itex]R^{4}[/itex] , 7![itex]R^{6}[/itex] which is (2n+1)! [itex]R^{2n}[/itex] however I need a bit of help condensing the numerator into a single series term.
 
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The Attempt at a SolutionI think it can be condensed into:-\frac{2k}{w_{o}^{2}} \sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n+1)!} (\frac{k}{w_{o}^{2}})^{n-1}But I'm not sure.
 

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