Comparison Tests for Series

Held under a convergent series

Comparison judges a series of non-negative terms against a simpler series whose behavior is already known, usually a p-series or a geometric series.

Suppose $0 \le a_{n} \le b_{n}$ for every n, and $\sum b_{n}$ converges to B. The partial sums of $\sum a_{n}$ climb, since no term is negative, and each one is at most the matching partial sum of $\sum b_{n}$, which is at most B. A total that climbs and never passes B settles on a limit, so $\sum a_{n}$ converges, to at most B.

Held above a divergent series

The other direction: if $0 \le b_{n} \le a_{n}$ for every n and $\sum b_{n}$ diverges, then $\sum a_{n}$ diverges. The partial sums of $\sum b_{n}$ grow past every number, and those of $\sum a_{n}$ are at least as large.

So the inequality must point the right way for the verdict wanted. Below a convergent series proves convergence. Above a divergent series proves divergence.

An example of each

$\sum \frac{1}{{n^{2} + 1}}$ converges. Its denominator is bigger than $n^{2}$, so each term is smaller than $\frac{1}{n^{2}}$, and $\sum \frac{1}{n^{2}}$ is a p-series with p = 2, which converges. The sum is less than $\frac{\pi^{2}}{6}$, about 1.645; it is about 1.077.

$\sum \frac{1}{\sqrt{n}}$ diverges. For $n \ge 1$, $\sqrt{n} \le n$, so $\frac{1}{\sqrt{n}} \ge \frac{1}{n}$, and the harmonic series $\sum \frac{1}{n}$ diverges.

The terms $\frac{1}{{n^{2} + 1}}$ for n = 1 to 6, with the vertical scale stretched: $\frac{1}{2}$, $\frac{1}{5}$, $\frac{1}{10}$, $\frac{1}{17}$, $\frac{1}{26}$, $\frac{1}{37}$. Each lies under the gold curve $y = \frac{1}{x^{2}}$, at the matching term of $\sum \frac{1}{n^{2}}$.

When the inequality points the wrong way

Take $\sum \frac{1}{{2n - 1}} = 1 + \frac{1}{3} + \frac{1}{5} + \dots$. Since $2n - 1 \ge n$, each term is at most $\frac{1}{n}$. But being smaller than a divergent series proves nothing: $\sum \frac{1}{n^{2}}$ is smaller still and converges. That comparison is the wrong way round.

Choose the comparison the other way. Since 2n − 1 < 2n, each term is more than $\frac{1}{{2n}}$. And $\sum \frac{1}{{2n}}$ is half the harmonic series, which diverges. Now the inequality points the right way, and $\sum \frac{1}{{2n - 1}}$ diverges.

Sometimes no simple inequality points the right way. Take $\sum \frac{1}{{n^{2} - n + 1}} = 1 + \frac{1}{3} + \frac{1}{7} + \dots$. Its terms behave like $\frac{1}{n^{2}}$, but since $n^{2} - n + 1 \le n^{2}$, each is at least $\frac{1}{n^{2}}$. Being bigger than a convergent series proves nothing.

The limit comparison test

Compare the sizes of the terms instead, by their ratio. If $a_{n}$ and $b_{n}$ are positive and $\frac{a_{n}}{b_{n}}$ tends to a limit L with $0 < L < \infty$, then $\sum a_{n}$ and $\sum b_{n}$ both converge or both diverge.

The reason: for large n the ratio is close to L, so it lies between $\frac{L}{2}$ and 2L. Then $a_{n}$ lies between $\left(\frac{L}{2}\right)b_{n}$ and $2L \times b_{n}$. If $\sum b_{n}$ converges, so does the series with terms $2L \times b_{n}$, and $\sum a_{n}$ is held under it. If $\sum b_{n}$ diverges, so does the series with terms $\left(\frac{L}{2}\right)b_{n}$, and $\sum a_{n}$ is held above it. Direct comparison does the rest.

For $\sum \frac{1}{{n^{2} - n + 1}}$ against $\frac{1}{n^{2}}$, the ratio is $\frac{n^{2}}{{n^{2} - n + 1}} = \frac{1}{{1 - \frac{1}{n} + \frac{1}{n^{2}}}}$, which tends to 1. So the series converges, like $\sum \frac{1}{n^{2}}$.

A limit of 0 or infinity decides nothing by itself: it means the terms are of different sizes, and the comparison series was the wrong choice.

Choosing what to compare with

Keep the highest power of n in the top and in the bottom. In $\frac{{2n + 1}}{{n^{3} + 5}}$ the top behaves like 2n and the bottom like $n^{3}$, so the term behaves like $\frac{2n}{n^{3}} = \frac{2}{n^{2}}$. Compare with $\frac{1}{n^{2}}$.

The ratio is $\frac{{2n + 1}}{{n^{3} + 5}} \times n^{2} = \frac{{2n^{3} + n^{2}}}{{n^{3} + 5}}$. Dividing the top and the bottom by $n^{3}$ gives $\frac{{2 + \frac{1}{n}}}{{1 + \frac{5}{n^{3}}}}$, which tends to 2. Since $0 < 2 < \infty$ and $\sum \frac{1}{n^{2}}$ converges, $\sum \frac{{2n + 1}}{{n^{3} + 5}}$ converges.

The same way, $\frac{{n + 3}}{{n^{2} + 1}}$ behaves like $\frac{n}{n^{2}} = \frac{1}{n}$. The ratio against $\frac{1}{n}$ is $\frac{{n^{2} + 3n}}{{n^{2} + 1}}$, which tends to 1, and $\sum \frac{1}{n}$ diverges, so $\sum \frac{{n + 3}}{{n^{2} + 1}}$ diverges.

The ratio $\frac{{2n + 1}}{{n^{3} + 5}} \div \frac{1}{n^{2}}$ for n = 1 to 10: 0.5, 1.54, 1.97, then just over 2, highest at about 2.12 when n = 6, and falling back toward the gold line at 2.

The usual mistakes

Comparing in the wrong direction: showing a series is smaller than a divergent one, or bigger than a convergent one. Neither proves anything.

Using direct comparison with negative terms. Each term of −1 − 1 − 1 − … is less than $\frac{1}{n^{2}}$, and the series diverges. The test needs $0 \le a_{n}$.

Comparing with the wrong power. Against $\frac{1}{n^{3}}$, the ratio for $\frac{{2n + 1}}{{n^{3} + 5}}$ tends to infinity; against $\frac{1}{n}$, it tends to 0. Neither decides.

Reading a limit of 0 or infinity as a verdict. Only a finite, non-zero L lets the two series share their behavior.

Two aerials

In the application below, one aerial's harmonics are held under three times $\sum \frac{1}{n^{2}}$, and the other's are held above the harmonic series, so one total is finite and the other is not.