anonymous
  • anonymous
Does the series converge or diverge? Sum (1)/(n+sqrt(n)*ln(n)), n-> 1 to infinity. I guess its answer is found using limit comparison? I'm just not really sure how to apply it.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\sum _n^{\infty } \frac{1}{n+\sqrt{n} \text{Log}[n]} \] Mathematica reports: 1. Sum does not converge 2. The ratio test is inconclusive 3.The root test is inconclusive Browse over to WolframAlpha.com and paste in the following: Sum [1/(n + Sqrt[n]*Log[n]), {n, Infinity}]
anonymous
  • anonymous
compare your series to the p series 1/n , where p is 1 and therefore divergent. then we can use limit comparison test lim n-->infinity (the original series)/(1/n) , with this we get that the limit is 1 and thus we know that since the series 1/n diverges the original series will also diverge
anonymous
  • anonymous
Thank you, that helps.

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