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$0 ≤ aₙ ≤ bₙ for every n, and $\sum b_n$Σbₙ converges to B. The partial sums of $\sum a_n$Σaₙ climb, since no term is negative, and each one is at most the matching partial sum of $\sum b_n$Σbₙ, which is at most B. A total that climbs and never passes B settles on a limit, so $\sum a_n$Σaₙ converges, to at most B.

Held above a divergent series

The other direction: if $0\le b_n\le a_n$0 ≤ bₙ ≤ aₙ for every n and $\sum b_n$Σbₙ diverges, then $\sum a_n$Σaₙ diverges. The partial sums of $\sum b_n$Σbₙ grow past every number, and those of $\sum a_n$Σaₙ 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

$\sum \frac{1}{n^2+1}$Σ 1/(n² + 1) converges. Its denominator is bigger than $n^2$n², so each term is smaller than $\frac{1}{n^2}$1/n², and $\sum \frac{1}{n^2}$Σ 1/n² is a p-series with p = 2, which converges. The sum is less than $\frac{{\pi }^2}{6}$π²/6, about 1.645; it is about 1.077.

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

The terms $\frac{1}{n^2+1}$1/(n² + 1) for n = 1 to 6, with the vertical scale stretched: $\frac{1}{2}$1/2, $\frac{1}{5}$1/5, $\frac{1}{10}$1/10, $\frac{1}{17}$1/17, $\frac{1}{26}$1/26, $\frac{1}{37}$1/37. Each lies under the gold curve $y=\frac{1}{x^2}$y = 1/x², at the matching term of $\sum \frac{1}{n^2}$Σ 1/n².

When the inequality points the wrong way

Take $\sum \frac{1}{2n-1}=1+\frac{1}{3}+\frac{1}{5}+\dots$Σ 1/(2n − 1) = 1 + 1/3 + 1/5 + …. Since $2n-1\ge n$2n − 1 ≥ n, each term is at most $\frac{1}{n}$1/n. But being smaller than a divergent series proves nothing: $\sum \frac{1}{n^2}$Σ 1/n² 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}$1/(2n). And $\sum \frac{1}{2n}$Σ 1/(2n) is half the harmonic series, which diverges. Now the inequality points the right way, and $\sum \frac{1}{2n-1}$Σ 1/(2n − 1) diverges.

Sometimes no simple inequality points the right way. Take $\sum \frac{1}{n^2-n+1}=1+\frac{1}{3}+\frac{1}{7}+\dots$Σ 1/(n² − n + 1) = 1 + 1/3 + 1/7 + …. Its terms behave like $\frac{1}{n^2}$1/n², but since $n^2-n+1\le n^2$n² − n + 1 ≤ n², each is at least $\frac{1}{n^2}$1/n². Being bigger than a convergent series proves nothing.

The limit comparison test

Compare the sizes of the terms instead, by their ratio. If $a_n$aₙ and $b_n$bₙ are positive and $\frac{a_n}{b_n}$aₙ/bₙ tends to a limit L with $0<L<\infty$0 < L < ∞, then $\sum a_n$Σaₙ and $\sum b_n$Σbₙ both converge or both diverge.

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

For $\sum \frac{1}{n^2-n+1}$Σ 1/(n² − n + 1) against $\frac{1}{n^2}$1/n², the ratio is $\frac{n^2}{n^2-n+1}=\frac{1}{1-\frac{1}{n}+\frac{1}{n^2}}$n²/(n² − n + 1) = 1/(1 − 1/n + 1/n²), which tends to 1. So the series converges, like $\sum \frac{1}{n^2}$Σ 1/n².

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 $\frac{2n+1}{n^3+5}$(2n + 1)/(n³ + 5) the top behaves like 2n and the bottom like $n^3$n³, so the term behaves like $\frac{2n}{n^3}=\frac{2}{n^2}$2n/n³ = 2/n². Compare with $\frac{1}{n^2}$1/n².

The ratio is $\frac{2n+1}{n^3+5}\times n^2=\frac{2n^3+n^2}{n^3+5}$(2n + 1)/(n³ + 5) × n² = (2n³ + n²)/(n³ + 5). Dividing the top and the bottom by $n^3$n³ gives $\frac{2+\frac{1}{n}}{1+\frac{5}{n^3}}$(2 + 1/n)/(1 + 5/n³), which tends to 2. Since $0<2<\infty$0 < 2 < ∞ and $\sum \frac{1}{n^2}$Σ 1/n² converges, $\sum \frac{2n+1}{n^3+5}$Σ (2n + 1)/(n³ + 5) converges.

The same way, $\frac{n+3}{n^2+1}$(n + 3)/(n² + 1) behaves like $\frac{n}{n^2}=\frac{1}{n}$n/n² = 1/n. The ratio against $\frac{1}{n}$1/n is $\frac{n^2+3n}{n^2+1}$(n² + 3n)/(n² + 1), which tends to 1, and $\sum \frac{1}{n}$Σ 1/n diverges, so $\sum \frac{n+3}{n^2+1}$Σ (n + 3)/(n² + 1) diverges.

The ratio $\frac{2n+1}{n^3+5}÷\frac{1}{n^2}$(2n + 1)/(n³ + 5) ÷ 1/n² 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}$1/n², and the series diverges. The test needs $0\le a_n$0 ≤ aₙ.

Comparing with the wrong power. Against $\frac{1}{n^3}$1/n³, the ratio for $\frac{2n+1}{n^3+5}$(2n + 1)/(n³ + 5) tends to infinity; against $\frac{1}{n}$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 $\sum \frac{1}{n^2}$Σ 1/n², and the other's are held above the harmonic series, so one total is finite and the other is not.