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Use the Comparison Test to determine whether the series is convergent of divergent. sum of sqrt(n)/(n-1) from n=2 to infinite
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\[\sum_{n=2}^{\infty} \frac{ \sqrt{n} }{ n-1 }\]
evaluate some values of n (ex: n = 1, 2, 3, 4, ...) If, when evaluated, the values seems to be approaching a certain number, then it is convergent. if not, then it is diveregent
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