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 $α = \underset{n \to \infty}{lim sup} | a_{n} |^{\frac{1}{n}}$ .

If $α < 1$ , then the series $\sum_{n = m}^{\infty} a_{n}$ is absolutely convergent (and hence conditionally convergent).
If $α > 1$ , then the series $\sum_{n = m}^{\infty} a_{n}$ is not conditionally convergent (and hence is not absolutely convergent either).
If $α = 1$ , 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 $| a_{n} | < α^{n} for some α < 1$ . We know that the geometric series $\sum_{n = N}^{\infty} α^{n}$ 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 $α < 1$ .

We can find a $ε > 0$ such that $α + ε < 1$ .
There must exist an $N \geq m$ such that $| a_{n} |^{\frac{1}{n}} < α + ε$ (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: $| a_{n} | < (α + ε)^{n}$ for all $n \geq N$ .
The geometric series $\sum_{n = m}^{\infty} (α + ε)^{n}$ converges as $(α + ε) < 1$ .
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 $α > 1$ .

For any $N \geq m$ 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 $ε$ -close for all $ε > 0$ (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