(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[tex]\sum\limits_{n = 2}^\infty {\frac{{\sqrt n }}{{n - 1}}} [/tex]

2. Relevant equations

3. The attempt at a solution

[tex]\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}[/tex]

so it is divergent.

But in this example from the notes, the teacher used theLimit Comparison Test, taking the [tex]\mathop {\lim }\limits_{n \to \infty } \frac{{a_n }}{{b_n }}[/tex], and using this to justify that since b_{n}is is divergent, therefore a_{n}is 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

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Comparison test for convergence

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**