Nth Root Test

Theorem
Let $\displaystyle \sum_{n=1}^\infty a_n$ be a series in $\R$.

Let the sequence $\left \langle {a_n} \right \rangle$ be such that $\displaystyle \limsup_{n \to \infty} \left|{a_n}\right|^{1/n} = l$.

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

Absolute Convergence
Let $l < 1$.

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

Consider the real sequence $\langle {b_n} \rangle$ defined by $\langle {b_n} \rangle = \langle \left \vert {a_n} \right \vert \rangle$.

Then $\displaystyle l = \limsup_{n\to\infty} b_n^{1/n}$.

It follows from Terms of Bounded Sequence Within Bounds that for sufficiently large $n$, $b_n < \left({l + \epsilon}\right)^n$.

The series $\displaystyle \sum_{n=1}^\infty \left( l + \epsilon \right)^n$ converges by Sum of Infinite Geometric Progression.

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

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

Divergence
Let $l > 1$.

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

We prove, by contradiction, that there exist arbitrarily large $n$ such that $\left\vert a_n \right\vert > \left( l - \epsilon \right)^n$.

Suppose that there exist a finite upper bound for the set $\left\{n \in \N : \left\vert {a_n} \right\vert^{1/n} > l - \epsilon \right\}$.

Then for all sufficiently large $n$, $\left\vert {a_n} \right\vert^{1/n} \le l - \epsilon$.

However, this implies that $\displaystyle \limsup_{n\to\infty} \left\vert {a_n} \right\vert^{1/n} \le l - \epsilon$, which is false by the definition of $l$.

Therefore, there exist arbitrarily large $n$ such that $\left\vert {a_n} \right\vert > \left( l - \epsilon \right)^n$.

Thus $\displaystyle \lim_{n\to\infty}\left\vert {a_n} \right\vert \neq 0$, and so $\displaystyle \lim_{n\to\infty} a_n \neq 0$.

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