Nth Root Test

Theorem
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be a series of real numbers $\R$ or complex numbers $\C$.

Let the sequence $\sequence {a_n}$ be such that the limit superior $\ds \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} = l$.

Then:


 * If $l > 1$, the series $\ds \sum_{n \mathop = 1}^\infty a_n$ diverges.
 * If $l < 1$, the series $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely.

Absolute Convergence
Let $l < 1$.

Then let us choose $\epsilon > 0$ such that $l + \epsilon < 1$.

Consider the real sequence $\sequence {b_n}$ defined by $\sequence {b_n} = \sequence {\size {a_n} }$.

Here, $\size {a_n}$ denotes either the absolute value of $a_n$, or the complex modulus of $a_n$.

Then:
 * $\ds l = \limsup_{n \mathop \to \infty} {b_n}^{1/n}$

It follows from Terms of Bounded Sequence Within Bounds that for sufficiently large $n$,:
 * $b_n < \paren {l + \epsilon}^n$

By Sum of Infinite Geometric Sequence, the series $\ds \sum_{n \mathop = 1}^\infty \paren {l + \epsilon}^n$ converges.

By the comparison test, $\ds \sum_{n \mathop = 1}^\infty b_n$ converges.

Hence $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely by the definition of absolute convergence.

Divergence
Let $l > 1$.

Then we choose $\epsilon > 0$ such that $l - \epsilon > 1$.

that there exist an upper bound for the set:
 * $S := \set {n \in \N: \size {a_n}^{1/n} > l - \epsilon}$

Then for all sufficiently large $n$:
 * $\size {a_n}^{1/n} \le l - \epsilon$

However, this implies that:
 * $\ds \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} \le l - \epsilon$

which is false by the definition of $l$.

The set $S$, then, is not bounded.

This means that there exist arbitrarily large $n$ such that:
 * $\size {a_n} > \paren {l - \epsilon}^n$

Thus:
 * $\ds \lim_{n \mathop \to \infty} \size {a_n} \ne 0$

and so $\ds \lim_{n \mathop \to \infty} a_n \ne 0$.

Hence from Terms in Convergent Series Converge to Zero, $\ds \sum_{n \mathop = 1}^\infty a_n$ must be divergent.