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Ratio Test

Consider a series ∑n=1∞an\sum\limits^\infty_{n=1}a_n.
  • If lim⁡n→∞∣an+1an∣<1{\displaystyle \lim_{n\to\infty}}|\frac{a_{n+1}}{a_n}|<1, then ∑n=1∞an\sum\limits^\infty_{n=1}a_n converges (infact, it converges absolutely)
  • If lim⁡n→∞∣an+1an∣>1{\displaystyle \lim_{n\to\infty}}|\frac{a_{n+1}}{a_n}|>1, then ∑n=1∞an\sum\limits^\infty_{n=1}a_n diverges
  • If lim⁡n→∞∣an+1an∣=1{\displaystyle \lim_{n\to\infty}}|\frac{a_{n+1}}{a_n}|=1, then the test fails and we don't know whether the series converges or not

Identifying Clues

  • the terms look like a ratio of two functions (rational function)
  • the terms contain factorials(!) or powers

Practice Question

Determine whether ∑n=1∞(n2)!nn\sum\limits^\infty_{n=1}\frac{(n^2)!}{n^n} converges or diverges.

Practice Question

Determine whether the series ∑n=1∞1n!{\displaystyle \sum_{n=1}^\infty}\frac{1}{n!} converges or diverges.

Practice Question

The series ∑n=1∞an{\displaystyle \sum_{n=1}^\infty}a_n has terms that are defined recursively by a1=1a_{_1}=1 and an=2n3−14+n3⋅an−1a_n=\frac{2\sqrt{n^3}-1}{\sqrt{4+n^3}}\cdot a_{n-1}.
Determine whether the series converges or diverges.

Extra Practice