Calc6.20

From Example Problems
Jump to: navigation, search

Discuss the convergence or divergence of the series \sum _{{n=1}}^{{\infty }}{\frac  {\ln n}{n-3}}.

We know the series \sum _{{n=1}}^{{\infty }}{\frac  {1}{n}} diverges because it is a p-series with p=1. Since

{\frac  {1}{n}}<{\frac  {1}{n-3}}<{\frac  {\ln n}{n-3}}\,

so, by the direct comparison, our series diverges. We can use the direct comparison test because our series has positive terms and so does the series \sum _{{n=1}}^{{\infty }}{\frac  {1}{n}}. Again, this makes sense. If we have a series with positive terms which diverges, then it goes off to infinity. Another series with even bigger positive terms will clearly also go off to infinity.


Main Page : Calculus