- #1
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Homework Statement
\sum\limits_{n = 2}^\infty {\frac{{\sqrt n }}{{n - 1}}}
Homework Equations
The Attempt at a Solution
\begin{array}{l}<br /> a_n = \frac{{\sqrt n }}{{n - 1}},\,\,b_n = \frac{{\sqrt n }}{n} = \frac{{n^{1/2} }}{n} = \frac{1}{{n^{1/2} }}\,\,{\rm{is_ a_ p - series_ with }}p{\rm{ = }}\frac{1}{2},\,\frac{1}{2} < 1\,{\rm{diverges}} \\ <br /> \\ <br /> \frac{{\sqrt n }}{{n - 1}} > \frac{{\sqrt n }}{n}\, \\ <br /> \\ <br /> \end{array}
so it is divergent.
But in this example from the notes, the teacher used the Limit Comparison Test, taking the \mathop {\lim }\limits_{n \to \infty } \frac{{a_n }}{{b_n }}, and using this to justify that since bnis is divergent, therefore anis also divergent. I understand this logic, but why was the extra step necessary when the comparison test alone showed it to be divergent?
