# The Limit Comparison Test V1

I'm currently learning Calculus 2, more specifically I'm learning about sequences and series. I'm not enjoying this section as much as I thought I would, this is because I'm having to learn all these different tests to determine the convergence and being shown no justification as to why it works. I've been shown the proofs, but the proofs are not justifying to my mind why they even work.

Limit Comparison Test:

Suppose that we have two series $\displaystyle\sum a_n$ and $\displaystyle\sum b_n$ with $a_n\geq0$,$b_n>0$ $\forall n$. Define, $$c = \displaystyle\lim_{n\to \infty} \frac{a_n}{b_n}$$ if $c$ is positive (i.e. $c>0$) and is finite (i.e. $c<\infty$) then either both series converge or both series diverge.

The first question I'd like to ask is, why does this even work?

The second question I'd like to ask is, under what conditions does this work? Why do I ask this? Consider the two following series: $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^3}$ and $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$.Both are p-series and a p-series converges when $p>1$ and diverges when $p\leq1$. Therefore, the two series above converges. Trying to verify this with the limit comparison test would go something like this $$\displaystyle\lim_{n\to \infty} \frac{n^2}{n^3} = \displaystyle\lim_{n\to \infty} \frac{1}{n} =0$$ $c\not>0$ which is implying that both series don't converge. So, what is going on?

One small final question, I'd really like to improve in this part of the course and be in a position where I don't have to remember all these annoying tests and just be able to derive certain things from logic. Would this be too much to hope for considering I'm only doing a Calculus course and not something like Real-Analysis.

• read the proof to understand why it works Jan 9 '17 at 1:05
• That's not how the limit comparison test works. It says that if $\sum a_n$ converges, then $\sum b_n$ converges. If $\sum a_n$ diverges, then $\sum b_n$ diverges. And only if it holds for that limit. It doesn't so limit comparison test doesn't work there. Jan 9 '17 at 1:07
• By the way, here is the proof Jan 9 '17 at 1:31
• @SimpleArt Thanks, but I've seen that proof. Unfortunately, I'm still not able to conclude why it works even after looking at the proof :( Jan 9 '17 at 1:33

The limit comparison test is very powerful. One of my favorite applications is this: Determine the convergence/divergence of $$\sum_{n=1}^\infty \frac 1{n^{1+1/n}}.$$
The heuristic is very simple. For large $n$, we're saying that basically $a_n = cb_n$ (for some positive number $c$). Ignoring small values of $n$, then $\sum a_n = c\sum b_n$, so obviously both series converge together or diverge together. The rigorous proof is only slightly more intricate, sandwiching $a_n$ between $c'b_n$ and $c''b_n$ for $0<c'<c<c''$.
Your logic in your second question is flawed. We're not saying things are "if and only if." We're saying that provided $\lim a_n/b_n = c$ and $0<c<\infty$, we can make an inference. We (ostensibly) know nothing if $c=0$ or $c=\infty$.
You should try to develop intuition based on this sort of comparison. Have a short list of series you know are convergent and divergent. Then try to say to yourself, "When $n$ is large, what do these terms look like?" (I.e., what are a convenient $b_n$ and what is the $c$?) Try this out with the one I gave at the outset.