Calculus · Grade 12 and undergraduate

The Ratio and Root Tests

Quick answer

The ratio test looks at L, the limit of |aₙ₊₁/aₙ|. If L < 1 the series converges absolutely, behaving like a geometric series with ratio L; if L > 1 it diverges; if L = 1 the test says nothing. The root test uses the limit of |aₙ|^(1/n) in the same way. They are the tests of choice for factorials and nth powers, and the ratio test is the main tool for finding where a power series converges.

What you'll learn

  • Apply the ratio test, simplifying factorials and powers
  • Apply the root test to nth powers
  • Explain why L < 1 means convergence and L > 1 divergence
  • Recognize when the tests are inconclusive

Comparing a series with itself

The comparison tests need a benchmark series. The ratio test finds one inside the series itself, by asking how each term compares with the one before.

The ratio test

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

  • If L<1L < 1, the series ∑an\sum a_n converges absolutely.
  • If L>1L > 1 (including L=∞L = \infty), the series diverges.
  • If L=1L = 1, the test gives no conclusion.

Factorials simplify well in the ratio: (n+1)!=(n+1)⋅n!(n + 1)! = (n + 1) \cdot n!, so (n+1)!n!=n+1\tfrac{(n + 1)!}{n!} = n + 1.

Why the ratio decides

If consecutive terms shrink by a factor near LL, then far out the series behaves like a geometric series with ratio LL. Choose a number rr between LL and 11. From some point on, each term is less than rr times the one before, so the tail is smaller than a geometric series with ratio r<1r < 1, which converges. If L>1L > 1, the terms eventually grow, and the series diverges by the nth-term test. The ratio test measures how fast the terms shrink against the yardstick of geometric series. When L=1L = 1, no geometric series is close enough to compare.

The root test

When the whole term is an nnth power, take the nnth root instead.

The root test

Let L=lim⁡n→∞∣an∣1/nL = \lim_{n \to \infty} \lvert a_n \rvert^{1/n}. If L<1L < 1 the series converges absolutely, if L>1L > 1 it diverges, and if L=1L = 1 the test gives no conclusion.

The reason is the same: ∣an∣1/n≈L\lvert a_n \rvert^{1/n} \approx L means ∣an∣≈Ln\lvert a_n \rvert \approx L^n, a geometric sequence.

Worked examples

Common mistakes

Practice problems

  1. Does ∑n=1∞n23n\displaystyle\sum_{n=1}^{\infty} \frac{n^2}{3^n} converge?

    Answer

    Yes

    Full solution

    (n+1)23n+1⋅3nn2=13(n+1n)2→13<1\tfrac{(n + 1)^2}{3^{n+1}} \cdot \tfrac{3^n}{n^2} = \tfrac{1}{3}\left(\tfrac{n + 1}{n}\right)^2 \to \tfrac{1}{3} < 1.

  2. Does ∑n=1∞2nn\displaystyle\sum_{n=1}^{\infty} \frac{2^n}{n} converge?

    Answer

    No

    Full solution

    2n+1n+1⋅n2n=2nn+1→2>1\tfrac{2^{n+1}}{n + 1} \cdot \tfrac{n}{2^n} = \tfrac{2n}{n + 1} \to 2 > 1.

  3. Does ∑n=1∞n!nn\displaystyle\sum_{n=1}^{\infty} \frac{n!}{n^n} converge?

    Answer

    Yes

    Full solution

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

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

    Answer

    No

    Full solution

    The ratio is (2n+2)(2n+1)(n+1)2=2(2n+1)n+1→4>1\tfrac{(2n + 2)(2n + 1)}{(n + 1)^2} = \tfrac{2(2n + 1)}{n + 1} \to 4 > 1.

  5. Does ∑n=0∞5nn!\displaystyle\sum_{n=0}^{\infty} \frac{5^n}{n!} converge?

    Answer

    Yes

    Full solution

    5n+1→0<1\tfrac{5}{n + 1} \to 0 < 1.

  6. Does ∑n=1∞(3nn+2)n\displaystyle\sum_{n=1}^{\infty} \left(\frac{3n}{n + 2}\right)^n converge?

    Answer

    No

    Full solution

    The root test gives 3nn+2→3>1\tfrac{3n}{n + 2} \to 3 > 1.

  7. Does ∑n=2∞1(ln⁡n)n\displaystyle\sum_{n=2}^{\infty} \frac{1}{(\ln n)^n} converge?

    Answer

    Yes

    Full solution

    The root test gives 1ln⁡n→0<1\tfrac{1}{\ln n} \to 0 < 1.

  8. What does the ratio test say about ∑n=1∞n3n4+1\displaystyle\sum_{n=1}^{\infty} \frac{n^3}{n^4 + 1}, and does the series converge?

    Answer

    L=1L = 1, no conclusion; the series diverges.

    Full solution

    The ratio of polynomial terms approaches 11. Compare with 1n\tfrac{1}{n} instead: n3/(n4+1)1/n=n4n4+1→1\tfrac{n^3/(n^4 + 1)}{1/n} = \tfrac{n^4}{n^4 + 1} \to 1, and the harmonic series diverges.

  9. Does ∑n=1∞(−1)nn4n\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n n}{4^n} converge absolutely?

    Answer

    Yes

    Full solution

    The ratio of absolute values is n+14n→14<1\tfrac{n + 1}{4n} \to \tfrac{1}{4} < 1.

  10. A student finds L=1L = 1 from the ratio test for ∑1n2\sum \tfrac{1}{n^2} and concludes that the series diverges. What went wrong?

    Hint

    What does the ratio test say when L=1L = 1?

    Answer

    L=1L = 1 gives no conclusion. The series converges, as a pp-series with p=2p = 2.

    Full solution

    The ratio test decides only when L<1L < 1 or L>1L > 1. At L=1L = 1 the terms shrink slower than any geometric series, and such series can converge or diverge.

Frequently asked questions

What does the ratio test say?

Let L be the limit of |aₙ₊₁/aₙ|. If L < 1 the series converges absolutely, if L > 1 it diverges, and if L = 1 the test is inconclusive.

Why is the ratio test inconclusive when L = 1?

Series with L = 1 can go either way. Both the sum of 1/n, which diverges, and the sum of 1/n², which converges, have L = 1.

When should I use the ratio test?

When the terms contain factorials or exponentials like 2ⁿ, which simplify neatly in the ratio of consecutive terms.

When should I use the root test?

When the whole term is raised to the nth power, such as (n/(2n + 1))ⁿ; the nth root removes the power.

How do I simplify (n + 1)!/n!?

(n + 1)! = (n + 1) · n!, so the ratio is n + 1.

What to learn next

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