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Math and science::Analysis::Tao::07. Series

The Root Test

Let \( \sum_{n=m}^{\infty}a_n \) be a series of real numbers and let \( \alpha = \limsup_{n \rightarrow \infty}|a_n|^{\frac{1}{n} } \).

  1. If \( \alpha < 1 \), then the series \( \sum_{n=m}^{\infty}a_n \) is absolutely convergent (and hence conditionally convergent).
  2. If \( \alpha > 1 \), then the series \( \sum_{n=m}^{\infty}a_n \) is not conditionally convergent (and hence is not absolutely convergent either).
  3. If \( \alpha = 1 \), this test does not assert any conclusion.

The famous Root Test.


Intuition

The root test applies the same principle used for the ratio test: that of comparing with a geometric series.

If \( |a_n|^{\frac{1}{n}} < 1 \) for all \( n \) greater than some \( N \), then we have \( |a_n| < \alpha^n \text{ for some } \alpha < 1 \). We know that the geometric series \( \sum_{n=N}^{\infty}\alpha^n \) converges, so by the comparison principle, then so too does the series \( \sum_{n=m}^{\infty}a_n \).

Proof

The proof follows the same lines as the intuition; added details cover the behaviour of the limit supremum.

First suppose that \( \alpha < 1 \).

  • We can find a \( \varepsilon > 0 \) such that \( \alpha + \varepsilon < 1 \).
  • There must exist an \( N \ge m \) such that \( |a_n|^{\frac{1}{n}} < \alpha + \varepsilon \) (By Prop 6.4.12(a): terms are eventually all less than any number larger than the limit supremum).
  • Manipulating the exponent we can write: \( |a_n| < (\alpha + \varepsilon)^n \) for all \( n \ge N \).
  • The geometric series \( \sum_{n=m}^{\infty} (\alpha + \varepsilon)^n \) converges as \( (\alpha + \varepsilon) < 1 \).
  • By the comparison principle, \( \sum_{n=m}^{\infty} |a_n| \) must also converge.
  • Thus, \( \sum_{n=m}^{\infty} a_n \) is absolutely convergent.

Now suppose that \( \alpha > 1 \).

  • For any \( N \ge m \) there exists a \( k \) such that \( |a_k|^{\frac{1}{k}} > 1 \) (By Prop 6.4.12(b): there are always some terms greater than any number lower than the limit supremum).
  • For such \( k \), we also have \( |a_k| > 1 \).
  • Thus, \( (a_n)_{n=m}^{\infty} \) is not eventually \( \varepsilon \)-close for all \( \varepsilon > 0 \) (e.g. 1-close, 0.5-close).
  • By the Zero Test we can conclude that \( \sum_{n=m}^{\infty} a_n \) is not conditionally convergent.

Source

p179