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Mathematics 22 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

Diverge, because n! has a higher order than b^n for any constant b.

OpenStudy (mathteacher1729):

Do you mean \[\frac{n!}{(n-1)3^n}\] or \[\frac{n!}{(n-1)^{3n}}\] ?

OpenStudy (mathteacher1729):

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."

OpenStudy (anonymous):

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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