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Infinite Series

Ratio Test

Definition

Definition of Ratio Test

Let ∑n=1∞an\sum_{n=1}^{\infty} a_n be a positive-term series with an>0a_n > 0. If:

lim⁡n→∞an+1an=ρ\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \rho

Then:

  1. If ρ<1\rho < 1, the series converges (convergence)
  2. If ρ>1\rho > 1, the series diverges (divergence)
  3. If ρ=1\rho = 1, the test is inconclusive
符号说明
SymbolTypePronunciation/ExplanationMeaning in This Article
ρ\rhoGreek letterRhoRepresents the limit value in series convergence tests
∑\sumGreek letterSigmaSummation symbol, representing series
∞\inftyMathematical symbolInfinityRepresents infinite series, infinite number of terms
lim⁡\limMathematical symbolLimitRepresents limit of sequence or function

Formula

Ratio Test Formula

lim⁡n→∞an+1an=ρ\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \rho

  • If ρ<1\rho < 1, the series converges
  • If ρ>1\rho > 1, the series diverges
  • If ρ=1\rho = 1, the test is inconclusive

Applicable Cases

  • General terms containing factorials, powers, etc.
  • Ratio of consecutive terms is easy to calculate

Examples

Example 1

Determine the convergence of the series ∑n=1∞n!nn\sum_{n=1}^{\infty} \frac{n!}{n^n}.

Solution: an=n!nna_n = \frac{n!}{n^n}

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

lim⁡n→∞an+1an=lim⁡n→∞(nn+1)n=1e<1\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^n = \frac{1}{e} < 1

Therefore, the series converges.

Exercises

Exercise 1

Determine the convergence of the series ∑n=1∞n2n\sum_{n=1}^{\infty} \frac{n}{2^n}.

Reference Answer (2 个标签)
series convergence ratio test

Problem-solving approach: Use the ratio test to compute the limit of the ratio between consecutive terms.

Detailed steps:

  1. Let an=n2na_n = \frac{n}{2^n}
  2. Calculate the ratio: an+1an=n+12n+1⋅2nn=n+12n\frac{a_{n+1}}{a_n} = \frac{n+1}{2^{n+1}} \cdot \frac{2^n}{n} = \frac{n+1}{2n}
  3. Find the limit: lim⁡n→∞an+1an=lim⁡n→∞n+12n=12<1\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{n+1}{2n} = \frac{1}{2} < 1
  4. Determine convergence: Ratio is less than 1, so the series converges

Answer: The series converges (convergence).

Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
eeMathematical symbolEuler’s numberBase of the natural logarithm, approximately 2.71828

Chinese-English Glossary

Chinese TermEnglish TermIPA PronunciationExplanation
比值判别法ratio test/ˈreɪʃiəʊ test/Method to determine series convergence using the ratio of consecutive terms
达朗贝尔判别法d’Alembert’s test/dælˈæmbəts test/Another name for the ratio test
正项级数positive series/ˈpɒzətɪv ˈsɪəriːz/Series where all terms are non-negative
收敛convergence/kənˈvɜːdʒəns/Sequence of partial sums has a finite limit
发散divergence/daɪˈvɜːdʒəns/Sequence of partial sums has no finite limit
PreviousComparison TestLearn the definition, usage, and applications of the comparison test. NextRoot TestLearn the definition, usage, and applications of the root test (Cauchy's test).

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