# Comparison test for convergence

1. Dec 9, 2007

### tony873004

1. The problem statement, all variables and given/known data
$$\sum\limits_{n = 2}^\infty {\frac{{\sqrt n }}{{n - 1}}}$$

2. Relevant equations

3. The attempt at a solution

$$\begin{array}{l} a_n = \frac{{\sqrt n }}{{n - 1}},\,\,b_n = \frac{{\sqrt n }}{n} = \frac{{n^{1/2} }}{n} = \frac{1}{{n^{1/2} }}\,\,{\rm{is_ a_ p - series_ with }}p{\rm{ = }}\frac{1}{2},\,\frac{1}{2} < 1\,{\rm{diverges}} \\ \\ \frac{{\sqrt n }}{{n - 1}} > \frac{{\sqrt n }}{n}\, \\ \\ \end{array}$$
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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution