# Proving divergence with the LCT

How do I do this? According to my book there are only three cases;

L > 0 means converges if one of the series converges

L = $\infty$ converges if one of the series converges

L = 0 if one of the series converges

This is confusing to me also because according to the proof it shouldn't matter what my $A_n$ or my $B_n$ is but according to this it does so I should say

L > 0 means $a_n$ converges if$b_n$ converges

L = $\infty$ if $a_n$ converges then $b_n$ converges

L = 0 if $b_n$ converges so does $a_n$

using

$\frac{a_n}{b_n}$

So how do I prove divergence? None of these even mention it.

The formulation stated here is incomplete, here is a more informative one:

Let $a_n,b_n>0$ be two sequences of real, positive, numbers, and consider the limit $$L=\lim_{n \to \infty} \frac{a_n}{b_n}$$

• If $L>0$, the series converge and diverge together (this means that $\sum a_n < \infty \Leftrightarrow \sum b_n< \infty$ and $\sum a_n=\infty \Leftrightarrow \sum b_n=\infty$)

• If $L=\infty$, $\sum b_n=\infty \implies \sum a_n=\infty$ and $\sum a_n< \infty \implies \sum b_n < \infty$

• If $L=0$, $\sum b_n< \infty \implies \sum a_n < \infty$ and $\sum a_n =\infty \implies \sum b_n=\infty$

I hope this helps.

• I guess I am not clever enough to see the connection to divergence. – Paul the Pirate Aug 13 '13 at 20:58
• @PaulthePirate This is called contraposition: if $A \implies B$ then $\not B \implies \not A$. – user1337 Aug 13 '13 at 21:09
• So how do I prove that something diverges? I need to find a $b_n$ that diverges? – Paul the Pirate Aug 13 '13 at 21:13
• @PaulthePirate it depends what $L$ is: if $L>0$ and one the series diverge, so is the other. If $L=\infty$ and $b_n$ diverges, so does $a_n$. If $L=0$ and $a_n$ diverges, so does $b_n$. – user1337 Aug 13 '13 at 21:21
• But say I have something like $\frac{1}{4n+ln(n)}$ obviously that diverges, but how do I prove that? – Paul the Pirate Aug 13 '13 at 21:27