The Limit Comparison Test

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The comparison test lets us compare an unknown series to a simpler, known series. But it requires us to come up with inequalities that can be difficult to figure out. Worse, sometimes it just doesn’t quite work. For example, suppose we’re looking at the series

    \[ \sum_{n=2}^{\infty}\frac{1}{n^2-1}. \]

This series looks almost like a p-series with p=2, so we expect it to converge. However, if we try to use the comparison test to show that, we end up comparing n^2-1 with n^2. We have

    \[ n^2-1 < n^2, \]

so

    \[ \frac{1}{n^2-1} > \frac{1}{n^2}. \]

But that’s not the inequality we want! Using this inequality we can only show that our series has terms that are larger than the corresponding terms of a converging series. That doesn’t say anything about whether or not our series converges. So how can we proceed?

Statement of the Test

There are a couple of ways to salvage our argument, but one of the simplest is to use the Limit Comparison Test. This test says that if \sum a_n and \sum b_n are series with positive terms and the limit

    \[ \lim_{n\rightarrow\infty}\frac{a_n}{b_n} \]

exists and is nonzero, then either both \sum a_n and \sum b_n converge, or else both diverge.

In our example above, we want to compare the series

    \[ \sum_{n=2}^{\infty}\frac{1}{n^2-1} \]

and

    \[ \sum_{n=2}^{\infty}\frac{1}{n^2}. \]

These series both have positive terms, so we only need to calculate the limit

    \begin{align*} \lim_{n\rightarrow\infty}\frac{\frac{1}{n^2-1}}{\frac{1}{n^2}} & = \lim_{n\rightarrow\infty}\frac{n^2}{n^2-1}\\ & = 1. \end{align*}

By the limit comparison, both series above either converge or diverge. Since we already know that the second one converges (because it’s a p-series with p>1), the first one must converge, too.

Series Whose Terms Are Quotients of Polynomials

The limit comparison test works great with series whose terms are quotients of polynomials. As an example, let’s take a look at the series

    \[ \sum_{i=1}^{\infty}\frac{4i^2+i+1}{i^3+i^2+2i+1}. \]

What are the largest terms in the numerator and denominator? In the numerator, 4i^2 is growing the fastest. In the denominator it’s i^3. So we expect the series to behave like the simpler series

    \[ \sum_{i=1}^{\infty}\frac{4i^2}{i^3} = 4\sum_{i=1}^{\infty}\frac{1}{n}. \]

This series is the harmonic series, so it diverges.

Both series have positive terms, so to use the limit comparison test we need to compute the limit

    \begin{align*} \lim_{i\rightarrow\infty}\frac{\frac{4i^2+i+1}{i^3+i^2+2i+1}}{\frac{4}{i}} & = \lim_{i\rightarrow\infty}\frac{4i^3+i^2+i}{4i^3+4i^2+8i+4} \\ & = 1. \end{align*}

Since the limit exists and isn’t zero, the limit comparison test tells us that both series must diverge.