# Limit Comparison Test

• Mar 28th 2011, 04:00 PM
eulcer
Limit Comparison Test
I've been teaching myself Calc. BC this year for the AP test, and this test for infinite series I've proved to be useful in analyzing these things, but I'm not sure how it was derived. I think I'm missing something simple, as the proof in my book is very short, but for this same reason maybe I haven't totally grasped it yet. My Calculus teacher hates infinite series (for whatever reasons) and therefore wasn't much help. Could someone try to break it down for me? Thanks in advance.

If An and Bn are both greater than zero
(lim n->infinity (An/Bn))>0
Then both series converge or diverge.
• Mar 28th 2011, 04:30 PM
roninpro
You should try to find a better teacher!

Suppose that $\lim_{n\to \infty} A_n/B_n=L>0$. This means that the sequence $\{A_n/B_n\}$ can be bounded above and below, say

$m

Multiplying both sides by $B_n$: $mB_n. Then we have

$m\sum B_n<\sum A_n

If $\sum A_n$ converges, then $\sum B_n<(1/m) \sum A_n$, so $\sum B_n$ converges by comparison. On the other hand, if $\sum B_n$ converges, then $\sum A_n, so $\sum A_n$ converges.

In summary, if the limit condition above holds, then $\sum A_n$ and $\sum B_n$ either both converge or diverge.

Let us know if you have any questions about this.
• Mar 28th 2011, 04:54 PM
eulcer
Quote:

Originally Posted by roninpro
You should try to find a better teacher!

Suppose that $\lim_{n\to \infty} A_n/B_n=L>0$. This means that the sequence $\{A_n/B_n\}$ can be bounded above and below, say

$m

Multiplying both sides by $B_n$: $mB_n. Then we have

$m\sum B_n<\sum A_n

If $\sum A_n$ converges, then $\sum B_n<(1/m) \sum A_n$, so $\sum B_n$ converges by comparison. On the other hand, if $\sum B_n$ converges, then $\sum A_n, so $\sum A_n$ converges.

In summary, if the limit condition above holds, then $\sum A_n$ and $\sum B_n$ either both converge or diverge.

Let us know if you have any questions about this.

I'm looking here for the teacher.

I get the "boundness" of the sequence, but going from sequence to series is what gets me.
Why can I say that the sums of the terms will be bounded accordingly too?

I get it, sorry.
If every term is greater or less, than the sum of course will be.
Thanks again.
I'm sure you'll see me here again before May.