Determine the convergence or divergence of the series. (2n)!/[(n-1)3^n]

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Determine the convergence or divergence of the series. (2n)!/[(n-1)3^n]

Mathematics
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Diverge, because n! has a higher order than b^n for any constant b.
Do you mean \[\frac{n!}{(n-1)3^n}\] or \[\frac{n!}{(n-1)^{3n}}\] ?
Either of those is sequence, not a series. A sequence effectively a list of numbers. A series tells you to "add up all the numbers in the sequence."

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Regardless of whether they're sequences or series, the nth terms don't converge anyway. I didn't notice the second interpretation. Hmm...

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