The Ratio Test

A fixed ratio

In a geometric series every term is the one before multiplied by the same number r, the common ratio. In $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots$1 + 1/2 + 1/4 + 1/8 + …, each term is half of the one before. A geometric series converges when |r| < 1 and diverges when $|r|\ge 1$|r| ≥ 1, and when it converges its sum is $\frac{a}{1-r}$a/(1 − r), which here is $\frac{1}{1-\frac{1}{2}}=2$1/(1 − 1/2) = 2.

Most series have no fixed ratio. Divide each term of $1+2+2+\frac{4}{3}+\frac{2}{3}+\dots$1 + 2 + 2 + 4/3 + 2/3 + … by the one before and the answers are 2, then 1, then $\frac{2}{3}$2/3, then $\frac{1}{2}$1/2: the ratio changes from term to term. The ratio test asks where that ratio settles as the terms go on, and treats the rest of the series as a geometric series with that ratio.

Each bar is a running total of the geometric series $1+r+r^2+\dots$1 + r + r² + …. At r = 0.8 the first four totals are 1, 1.8, 2.44 and 2.952, and the dashed line is the sum, $\frac{1}{1-0.8}=5$1/(1 − 0.8) = 5. Drag N to add terms: the totals climb toward 5 and never cross it. Drag r past 1 and they run off the top.

The test

Write the series as $a_1+a_2+a_3+\dots$a₁ + a₂ + a₃ + …, and divide each term by the one before it: $a_{n+1}/a_n$aₙ₊₁ / aₙ. Take its size, $|a_{n+1}/a_n|$|aₙ₊₁ / aₙ|, so that negative terms make no difference, and let L be the number it approaches as n grows.

If L < 1, the series converges. If L > 1, it diverges. If L = 1, the test decides nothing, and another test is needed.

Here is why L < 1 is enough. Choose a number r between L and 1. Once n is large enough, every ratio is below r, so from that term on each term is less than r times the one before in size. The sizes of the remaining terms are then smaller than the terms of a geometric series with ratio r, and that series converges because r < 1. So the sizes add up to a finite total. The series converges absolutely: it would converge even with every term made positive.

If L > 1, then once n is large enough every term is larger in size than the one before. The terms grow instead of shrinking toward 0, and a series whose terms do not tend to 0 diverges.

Factorials underneath

Take $\sum \frac{2^n}{n!}$Σ 2ⁿ/n!, starting from n = 0: $1+2+2+\frac{4}{3}+\frac{2}{3}+\dots$1 + 2 + 2 + 4/3 + 2/3 + …. To divide one term by the one before, use two facts: $2^{n+1}=2\times 2^n$2ⁿ⁺¹ = 2 × 2ⁿ, and (n + 1)! = (n + 1) × n!. So the next term is this term multiplied by 2 and divided by n + 1, and the ratio is $\frac{2}{n+1}$2/(n + 1).

As n grows, $\frac{2}{n+1}$2/(n + 1) tends to 0, so L = 0. That is less than 1, so the series converges. Each new term gains a factor of 2 on top and a factor of n + 1 underneath, and from n = 2 on the factor underneath is the larger one.

The running totals agree. Term by term they are 1, 3, 5, 6.333333, 7, 7.266667, 7.355556, 7.380952 and 7.387302, closing in on $e^2=7.389056$e² = 7.389056. This is the series for $e^x$eˣ with x = 2.

The terms $\frac{2^n}{n!}$2ⁿ/n! for n = 0 to 8. They rise from n = 0 to n = 1, where the ratio is 2, stay level from n = 1 to n = 2, where it is 1, and then fall faster and faster as the ratio $\frac{2}{n+1}$2/(n + 1) drops further below 1.

The same terms turned over

Turn every term over and the nth term becomes $n!÷2^n$n! ÷ 2ⁿ: $1+\frac{1}{2}+\frac{1}{2}+\frac{3}{4}+\frac{3}{2}+\frac{15}{4}+\dots$1 + 1/2 + 1/2 + 3/4 + 3/2 + 15/4 + …. Now the factorial is on top, so the next term is this one multiplied by n + 1 and divided by 2, and the ratio is $\frac{n+1}{2}$(n + 1)/2.

As n grows, $\frac{n+1}{2}$(n + 1)/2 grows without limit, so L is larger than 1 and the series diverges. From n = 2 on, the ratio is above 1 and every term is bigger than the one before; by n = 8 the term is 157.5. The terms do not tend to 0, which is the nth term test reaching the same verdict.

When L = 1

For the harmonic series $\sum \frac{1}{n}$Σ 1/n, each term divided by the one before is $\frac{n}{n+1}$n/(n + 1). At n = 10 that is 0.909091, at n = 100 it is 0.990099, and at n = 1000 it is 0.999001, so L = 1. For $\sum \frac{1}{n^2}$Σ 1/n² the ratio is $\frac{n^2}{{\left(n+1\right)}^2}$n²/(n + 1)², which is 0.826446 at n = 10, 0.980296 at n = 100 and 0.998003 at n = 1000, so L = 1 here too.

Yet the two series do opposite things. $\sum \frac{1}{n}$Σ 1/n diverges: its running total is 2.929 after 10 terms, 5.187 after 100 and 7.485 after 1000, and it keeps growing. $\sum \frac{1}{n^2}$Σ 1/n² converges, by the p-series test with p = 2: its running totals at the same points are 1.549768, 1.634984 and 1.643935, closing in on $\frac{{\pi }^2}{6}=1.644934$π²/6 = 1.644934.

So L = 1 cannot tell convergence from divergence, and another test has to decide: the p-series test, a comparison, or the alternating series test. The ratio test works best when the terms contain factorials or powers such as $2^n$2ⁿ. When the terms are built from powers of n alone, it gives L = 1.

The gold curve is the ratio $\frac{n}{n+1}$n/(n + 1) for $\sum \frac{1}{n}$Σ 1/n, and the plain one is $\frac{n^2}{{\left(n+1\right)}^2}$n²/(n + 1)² for $\sum \frac{1}{n^2}$Σ 1/n², plotted against n. Both climb toward the dashed line at 1 without reaching it, and one series diverges while the other converges.

The usual mistakes

Dividing the wrong way round. The test divides each term by the one before it. Dividing a term by the one after it gives the reciprocal ratio, so a limit of 0 becomes one that grows without limit, and the verdict flips.

Canceling the factorials wrongly. (n + 1)! divided by n! is n + 1, not n, because (n + 1)! = (n + 1) × n!.

Reading L = 1 as a verdict. $\sum \frac{1}{n}$Σ 1/n and $\sum \frac{1}{n^2}$Σ 1/n² both give L = 1; the first diverges and the second converges.

Judging by the first few ratios. The first ratio of $\sum \frac{2^n}{n!}$Σ 2ⁿ/n! is 2, which is above 1, and still the series converges. The test uses the limit of the ratio, not its early values.

Calls at a helpdesk

In the application below, the chance of n calls in one minute has n! underneath, the same shape as $\frac{2^n}{n!}$2ⁿ/n! with 3 in place of 2. The ratio test shows that the chances add up to a finite total, and the ratio $\frac{3}{n+1}$3/(n + 1) also shows where the chances are largest.