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sammii2uBest ResponseYou've already chosen the best response.0
\[\sum_{n=1}^{\infty} \frac{ e^{\sqrt{n}} }{ \sqrt{n} }\] \[\sum_{n=2}^{\infty} \frac{ 1 }{ n(\ln n)^3 }\]
 one year ago

geerky42Best ResponseYou've already chosen the best response.0
I don't do this thing yet so I may be wrong... Ask yourself a question. Which increase faster as n approaches to infinity? \(\large e^{\sqrt{n}}\) or \(\sqrt{n}\). If \(\large e^{\sqrt{n}}\) increase faster or approach to something greater than \(\sqrt{n}\), then it diverges. Otherwise it converges. What does \(n(\ln n )^3\) approach to when n approach to infinity? If it approaches to something under one, then it diverges, otherwise it converges.
 one year ago

amoodaryaBest ResponseYou've already chosen the best response.1
for the first use ratio test that is converegent and for the second use integral test see what i drawdw:1360527817600:dw that is also conv
 one year ago
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