Unit 7, lesson 101

The Ratio and Root Tests

How the ratio and root tests compare a series with a geometric series, when they prove absolute convergence or divergence, and why a limit of exactly one decides nothing.

  • Theory part 201
  • Practice part 202
  • Convergence and Convergence Tests

A geometric series converges when each term is a fixed fraction of the one before it. Many series are not geometric, but far out they behave almost like one. The ratio and root tests measure that limiting ratio. They are the natural tools for terms built from factorials, exponentials and nnth powers.

Key terms

Ratio test

Definition Let ∑an\sum a_n be a series with an≠0a_n \ne 0 for all n≥Nn \ge N, and suppose

L=lim⁡n→∞∣an+1an∣L = \lim_{n \to \infty} \left\lvert \frac{a_{n+1}}{a_n} \right\rvert

exists as a real number or equals ∞\infty.

  1. If L<1L < 1, the series converges absolutely.
  2. If L>1L > 1 or L=∞L = \infty, the series diverges.
  3. If L=1L = 1, the test gives no conclusion.

In plain words Far out, each term is roughly LL times the one before. If L<1L < 1 the series behaves like a convergent geometric series, and if L>1L > 1 the terms grow.

Example For ∑n22n\sum \frac{n^2}{2^n},

∣an+1an∣=(n+1)22n+1⋅2nn2=12(1+1n)2→12<1,\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{(n + 1)^2}{2^{n+1}} \cdot \frac{2^n}{n^2} = \frac{1}{2}\left(1 + \frac{1}{n}\right)^2 \to \frac{1}{2} < 1,

so the series converges absolutely.

Root test

Definition For a series ∑an\sum a_n suppose

L=lim⁡n→∞∣an∣nL = \lim_{n \to \infty} \sqrt[n]{\lvert a_n \rvert}

exists as a real number or equals ∞\infty.

  1. If L<1L < 1, the series converges absolutely.
  2. If L>1L > 1 or L=∞L = \infty, the series diverges.
  3. If L=1L = 1, the test gives no conclusion.

In plain words If ∣an∣\lvert a_n \rvert is roughly LnL^n for large nn, the series behaves like a geometric series with ratio LL. The root test reads off that LL directly.

Example For ∑(2n+33n+1)n\sum \left(\frac{2n + 3}{3n + 1}\right)^n the terms are positive and

ann=2n+33n+1→23<1,\sqrt[n]{a_n} = \frac{2n + 3}{3n + 1} \to \frac{2}{3} < 1,

so the series converges absolutely.

Absolute convergence from either test

Definition When the ratio or root test gives L<1L < 1, the conclusion is absolute convergence. The proof shows that ∣an∣≤Crn\lvert a_n \rvert \le C r^n for all large nn, for any fixed rr with L<r<1L < r < 1 and some constant CC, and then compares ∑∣an∣\sum \lvert a_n \rvert with the geometric series ∑Crn\sum C r^n.

In plain words Both tests look only at absolute values, so when they say yes they give the strong kind of convergence. Signs never need checking.

Example For ∑(−5)nn!\sum \frac{(-5)^n}{n!} the ratio is

∣an+1an∣=5n+1→0<1,\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{5}{n + 1} \to 0 < 1,

so the series converges absolutely, even though the first few terms grow in size (55, 12.512.5, 20.820.8, 26.026.0 for n=1n = 1 to 44).

Divergence from either test

Definition When the ratio or root test gives L>1L > 1 or L=∞L = \infty, the terms satisfy ∣an∣↛0\lvert a_n \rvert \not\to 0. For the ratio test, ∣an+1∣>∣an∣\lvert a_{n+1} \rvert > \lvert a_n \rvert for all large nn. For the root test, ∣an∣>1\lvert a_n \rvert > 1 for all large nn. In either case the series diverges by the nth term test.

In plain words A limit above one means the terms are not getting smaller, so there is no chance of a finite sum.

Example For ∑(−1)n2nn3\sum \frac{(-1)^n 2^n}{n^3},

∣an+1an∣=2(nn+1)3→2>1,\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = 2\left(\frac{n}{n + 1}\right)^3 \to 2 > 1,

so the series diverges. Its terms grow in size without bound, even though they alternate in sign.

The inconclusive value one

Definition If the limit in the ratio test or the root test equals 11, the series may converge absolutely, converge conditionally or diverge. Another test is needed.

In plain words A limit of exactly one means the terms are not shrinking at any geometric rate. That leaves the question open.

Example Both ∑1n\sum \frac{1}{n} and ∑1n2\sum \frac{1}{n^2} give L=1L = 1 in the ratio test, since nn+1→1\frac{n}{n + 1} \to 1 and n2(n+1)2→1\frac{n^2}{(n + 1)^2} \to 1. The first diverges and the second converges. The alternating harmonic series also gives L=1L = 1, and it converges conditionally. The two tests always give 11 on p-series and on rational functions of nn.

Factorials and powers

Definition These facts come up again and again with the ratio test. For integers n≥0n \ge 0,

(n+1)!n!=n+1,(2n+2)!(2n)!=(2n+1)(2n+2),lim⁡n→∞(1+1n)n=e.\frac{(n + 1)!}{n!} = n + 1, \qquad \frac{(2n + 2)!}{(2n)!} = (2n + 1)(2n + 2), \qquad \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e.

Also nn→1\sqrt[n]{n} \to 1 and cn→1\sqrt[n]{c} \to 1 for every constant c>0c > 0.

In plain words Most of a factorial cancels in a ratio. When nnn^n appears, expect the number ee.

Example For ∑n!nn\sum \frac{n!}{n^n},

an+1an=(n+1)!(n+1)n+1⋅nnn!=nn(n+1)n=1(1+1n)n→1e<1.\frac{a_{n+1}}{a_n} = \frac{(n + 1)!}{(n + 1)^{n+1}} \cdot \frac{n^n}{n!} = \frac{n^n}{(n + 1)^n} = \frac{1}{\left(1 + \frac{1}{n}\right)^n} \to \frac{1}{e} < 1.

The series converges. At n=10n = 10 the ratio is already about 0.3860.386, close to 1e≈0.368\frac{1}{e} \approx 0.368.

Check yourself

  1. Does ∑n=1∞n!(2n)!\sum_{n=1}^{\infty} \frac{n!}{(2n)!} converge?
Show the answer

Yes. Most of each factorial cancels, and the ratio test gives

an+1an=(n+1)!n!⋅(2n)!(2n+2)!=n+1(2n+1)(2n+2)=12(2n+1)→0<1.\frac{a_{n+1}}{a_n} = \frac{(n + 1)!}{n!} \cdot \frac{(2n)!}{(2n + 2)!} = \frac{n + 1}{(2n + 1)(2n + 2)} = \frac{1}{2(2n + 1)} \to 0 < 1.
  1. Does ∑n=1∞(1−1n)n2\sum_{n=1}^{\infty} \left(1 - \frac{1}{n}\right)^{n^2} converge?
Show the answer

Yes. The terms are nonnegative, and the root test gives

ann=(1−1n)n→1e<1.\sqrt[n]{a_n} = \left(1 - \frac{1}{n}\right)^n \to \frac{1}{e} < 1.
  1. What does the ratio test say about ∑n+1n3\sum \frac{n + 1}{n^3}?
Show the answer

Nothing, because the limit is 11. Limit comparison with 1n2\frac{1}{n^2} shows the series converges.

How this lesson connects

It also uses factorials and roots.