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I know you have to do the ratio test for this one but I don't know how to do it.

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- #1

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I know you have to do the ratio test for this one but I don't know how to do it.

- #2

Tom Mattson

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[tex]s_n=\frac{n!^{{2a}^n}}{(2n)!}[/tex]

Can you write down [itex]s_{n+1}[/itex]?

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im sry its (n!)^2(a^n) for the numerator

- #4

HallsofIvy

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[tex]\left[\frac{((n+1)!)^2a^{n+1}}{(2(n+1))!}\right]\left[\frac{(2n)!}{((n!)^2a^n}\right]= \left[\frac{(n+1)!}{n!}\right]^2\left[\frac{a^{n+1}}{a^n}\right]\left[\frac{(2n)!}{(2(n+1))!}][/tex]

[tex]= \frac{(n+1)(a)}{(2n+2)(2n+1)[/tex]

What is the limit of that as n goes to infinity? If that limit depends on a, then the series will converge only for values of a that make the limit less than 1.

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