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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.

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