The Root Test
Let \( \sum_{n=m}^{\infty}a_n \) be a series of real numbers and let \( \alpha = \limsup_{n \rightarrow \infty}|a_n|^{\frac{1}{n} } \).
- If \( \alpha < 1 \), then the series \( \sum_{n=m}^{\infty}a_n \) is absolutely convergent (and hence conditionally convergent).
- If \( \alpha > 1 \), then the series \( \sum_{n=m}^{\infty}a_n \) is not conditionally convergent (and hence is not absolutely convergent either).
- 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.