Comparison Tests

A comparison test is simply sizing up a series you don't know against one you do: if your pile of numbers is always smaller than a pile you know is finite, yours is finite too. It's good because most series you'll meet aren't exactly a geometric series or a p-series, but they look a lot like one. The line $0 \le a_n \le b_n$ and $\sum b_n$ converges $\implies \sum a_n$ converges looks like a lot, but it's just three pieces: $a_n$ (the series you're stuck on), $b_n$ (a series you already understand, usually a p-series or geometric) and $a_n \le b_n$ (yours is smaller, term by term). The limit version, $\frac{a_n}{b_n} \to c$, just says the two are roughly proportional far out. It's a lot of notation for something you'd do with your eyes, so bear with it.

So far you've got a small toolkit of series you fully understand: geometric series, p-series, the harmonic series. Most series you meet won't be any of those exactly, but they'll look like one of them. Something like $\frac{1}{n^2 + 5}$ is basically $\frac{1}{n^2}$ with a little extra in the bottom. This lesson turns "basically" into a proof.

Direct comparison (positive terms, $a_n \le b_n$):

• $\sum b_n$ converges $\implies$ $\sum a_n$ converges (smaller than finite).
• $\sum a_n$ diverges $\implies$ $\sum b_n$ diverges (bigger than infinite).

Why does the first one work? With positive terms, the partial sums of $\sum a_n$ only go up, and they always stay below the total of $\sum b_n$. That's an increasing sequence with a ceiling, so by the monotone convergence theorem it has to converge.

A common mistake. It's tempting to use the other two directions too, since they look just as natural. They tell you nothing. Being smaller than a divergent series proves nothing: $\frac{1}{n^2}$ is smaller than $\frac1n$, and it converges, but $\frac{1}{2n}$ is smaller too, and it diverges. Same story for being bigger than a convergent one. Make sure your inequality points the useful way before you lean on it.

Choosing the comparison. Keep the dominant parts of the top and bottom, since those are what matter when $n$ is huge. For $\dfrac{3n + 1}{n^3 - 2}$ the dominant behavior is $\dfrac{3n}{n^3} = \dfrac{3}{n^2}$, so compare with $\sum \frac{1}{n^2}$.

Limit comparison saves you from proving inequalities, which can get fiddly (try showing $\frac{3n+1}{n^3-2} \le \frac{C}{n^2}$ for every $n$). If $\dfrac{a_n}{b_n} \to c$ with $0 < c < \infty$, the terms are eventually about $c$ times each other, and multiplying a series by a constant can't change whether it converges, so the series share a fate. Edge cases: if $c = 0$ and $\sum b_n$ converges, so does $\sum a_n$; if $c = \infty$ and $\sum b_n$ diverges, so does $\sum a_n$.

Engineers bound sums they can't compute exactly. If every error term in an approximation is below the matching term of a convergent geometric series, the total error is guaranteed finite, and bounded by that geometric sum.