5.4 Comparison Tests

Big idea

You now own two benchmark families whose behavior is completely settled: geometric series and p-series. The comparison tests are the machinery that leverages those benchmarks against everything else. Instead of analyzing a complicated series directly, you find a simpler one that behaves the same way and transfer the verdict.

The direct comparison is the honest version of the idea. If every term of your series is at most the corresponding term of a convergent series with nonnegative terms, your partial sums are bounded above by a finite number, and since they increase they must converge. Turn the inequality around and the same reasoning runs the other way: if every term is at least the corresponding term of a divergent series, your partial sums are pushed past every ceiling. The two directions are not interchangeable, and half the errors in this section come from using a true inequality that points the wrong way.

That fragility is what the limit comparison test fixes. Most series you meet are rational-looking expressions whose behavior for large $n$ is governed by the leading powers, and the exact inequality against the leading behavior often goes the wrong way by a little. The limit comparison test ignores the direction of the inequality and asks only whether the two sequences of terms are proportional in the limit. If the ratio settles on a positive finite number, the two series must do the same thing.

The practical skill is choosing the benchmark. Look at your term, discard everything that does not matter for large $n$, and keep the dominant power of $n$ or the dominant exponential. That surviving expression is your comparison series, and it is almost always a p-series or a geometric series, which is exactly why the previous section mattered.

Decoder

For series with nonnegative terms, being termwise no larger than a convergent series forces convergence, and being termwise no smaller than a divergent series forces divergence.

The hypothesis “nonnegative terms” is not decoration. The whole argument runs through the fact that the partial sums of a nonnegative series form an increasing sequence, so boundedness alone settles convergence. Introduce sign changes and partial sums can move in both directions, boundedness no longer implies convergence, and the argument collapses. For series with mixed signs, compare the absolute values instead, which is the subject of the next section.

The two directions are the part to internalize. Pushing your series under a convergent one is informative; pushing it under a divergent one is not, because being smaller than something infinite says nothing. Likewise being bigger than something finite says nothing. If your inequality points the useless way, either find a different benchmark or switch to the limit comparison test.

The limit comparison test asks a weaker question and therefore answers more often. It needs both series to have positive terms and it needs the ratio $a_n/b_n$ to approach a number $L$ with $0 < L < \infty$. Under those conditions the terms are eventually within a constant factor of each other in both directions, and a constant factor cannot change a verdict. When $L$ is $0$ or infinite you get only one direction, so treat those cases as special rather than as failures.

Definitions and results

Direct comparison test, convergent direction. If $0 \le a_n \le b_n$ for all large $n$ and $\sum b_n$ converges, then $\sum a_n$ converges.

Direct comparison test, divergent direction. If $0 \le b_n \le a_n$ for all large $n$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.

Why it works. With nonnegative terms the partial sums increase. In the first case they are bounded above by the sum of $\sum b_n$, and an increasing bounded sequence converges. In the second they exceed the unbounded partial sums of $\sum b_n$.

Limit comparison test. Suppose $a_n > 0$ and $b_n > 0$ for all large $n$ and

$$ L = \lim_{n \to \infty}\frac{a_n}{b_n} $$

If $0 < L < \infty$, then $\sum a_n$ and $\sum b_n$ both converge or both diverge. If $L = 0$ and $\sum b_n$ converges, then $\sum a_n$ converges. If $L = \infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges. The remaining combinations give no information.

Choosing a benchmark. Keep the highest power of $n$ in the numerator and in the denominator and discard the rest. A term behaving like $\frac{n^{j}}{n^{k}}$ suggests the p-series with $p = k - j$. A term containing $r^n$ with $|r| < 1$ suggests a geometric series.

Reading the benchmark’s verdict. The p-series $\sum n^{-p}$ converges when $p > 1$ and diverges when $p \le 1$. The geometric series $\sum r^n$ converges when $|r| < 1$. These two results carry almost every comparison in practice.

Only finitely many exceptions matter. All hypotheses need hold only for $n$ beyond some index, because dropping finitely many terms changes a sum but never a verdict. So a term that is negative or badly behaved for small $n$ is not an obstacle.

What comparison cannot do. It gives a verdict, never a value. A convergent comparison bounds the sum but does not produce it, and an unlucky benchmark produces no conclusion at all rather than a wrong one.

Worked examples

A comparison that works in the convergent direction

Decide $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2 + 4n}$.

The terms are positive. Since $n^2 + 4n > n^2$ for every $n \ge 1$, taking reciprocals reverses the inequality:

$$ 0 < \frac{1}{n^2 + 4n} < \frac{1}{n^2} $$

The benchmark $\sum \frac{1}{n^2}$ is a p-series with $p = 2 > 1$, so it converges. The inequality points the useful way, so the original series converges by direct comparison. Adding $4n$ to the denominator only made the terms smaller, which is exactly what you want when aiming for convergence.

A comparison that works in the divergent direction

Decide $\displaystyle\sum_{n=2}^{\infty}\frac{1}{\sqrt{n} - 1}$.

For $n \ge 2$ the denominator is positive and smaller than $\sqrt n$, so

$$ \frac{1}{\sqrt n - 1} > \frac{1}{\sqrt n} > 0 $$

The benchmark $\sum \frac{1}{\sqrt n}$ is a p-series with $p = \frac12 \le 1$, so it diverges. Being termwise larger than a divergent series, the original series diverges too.

When the inequality points the wrong way

Decide $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n^2 - 3}$.

The natural benchmark is $\sum \frac{1}{n^2}$, which converges. But $n^2 - 3 < n^2$, so

$$ \frac{1}{n^2 - 3} > \frac{1}{n^2} $$

and being larger than a convergent series tells you nothing. Direct comparison is inconclusive here, and it stays inconclusive no matter how long you stare at it.

The limit comparison test does not care about direction. With $a_n = \frac{1}{n^2-3}$ and $b_n = \frac{1}{n^2}$, both positive for $n \ge 2$,

$$ L = \lim_{n \to \infty}\frac{1/(n^2-3)}{1/n^2} = \lim_{n \to \infty}\frac{n^2}{n^2 - 3} = \lim_{n \to \infty}\frac{1}{1 - 3/n^2} = 1 $$

Since $L = 1$ is positive and finite, the two series share a verdict, and the benchmark converges. Therefore $\sum \frac{1}{n^2-3}$ converges. The subtraction of $3$ was never going to matter, and the limit comparison test is the tool that says so rigorously.

Picking the benchmark off the dominant powers

Decide $\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt{n} + 2}{n^2 + n}$.

For large $n$ the numerator behaves like $\sqrt n = n^{1/2}$ and the denominator like $n^2$, so the term behaves like $n^{1/2 - 2} = n^{-3/2}$. Take $b_n = n^{-3/2}$, which is positive, and compute the ratio:

$$ \frac{a_n}{b_n} = \frac{\sqrt n + 2}{n^2 + n}\cdot n^{3/2} = \frac{n^{2} + 2n^{3/2}}{n^2 + n} = \frac{1 + 2n^{-1/2}}{1 + n^{-1}} \longrightarrow 1 $$

The limit is positive and finite. The benchmark is a p-series with $p = \frac32 > 1$, so it converges, and therefore the original series converges.

Sanity check the size of the answer. The terms are near $n^{-3/2}$ for large $n$, and $n^{-3/2}$ sits between $n^{-1}$ and $n^{-2}$; the verdict of convergence depends on being on the correct side of $p = 1$, and $\frac32$ clears it. Had the numerator been $n + 2$ instead, the exponent would have been $-1$ and the series would have diverged, so the square root is doing real work here.

Practice

One skill covering both tests. Expect to pick a benchmark from the dominant powers, to check the direction of a direct comparison before trusting it, and to switch to the limit comparison test when the inequality is unhelpful or awkward to establish.

Quiz

Five items on direct comparison in both directions, the limit comparison test and its hypotheses, and choosing an appropriate benchmark series.