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Mathematics 24 Online
OpenStudy (anonymous):

sum from n=2 -> infinity (sqrt(n))/(n-1)

OpenStudy (anonymous):

\[\sum_{n-2}^{\infty} \sqrt{n}/ (n-1)\]

OpenStudy (anonymous):

comparison test gives me \[1/\sqrt{n}\]

OpenStudy (anonymous):

p=.5<1 divergent

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