Kuzee
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
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}.

series convergenceratio test

Approach: Use the ratio test to compute the limit of the ratio between consecutive terms.

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
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