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.

Proof

 * Suppose $l < 1$.

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

For $N$ big enough, $\left|{a_n}\right| < \left({l + \epsilon}\right)^n$ from Terms of Bounded Sequence Within Bounds.

But $\left \langle {\left({l + \epsilon}\right)^n} \right \rangle$ converges from Sum of Infinite Geometric Progression.

The fact that $\displaystyle \sum_{n=1}^\infty |a_n|$ converges absolutely follows from the Comparison Test.


 * Now suppose $l > 1$.

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

Then from Limit of Subsequence of Bounded Sequence, we can find a subsequence $\left \langle {a_{n_r}} \right \rangle$ of $\left \langle {a_n} \right \rangle$ such that $\left|{a_{n_r}}\right| > \left({l - \epsilon}\right)^{n_r}$ as $r \to \infty$.

Thus the terms of $\displaystyle \sum_{n=1}^\infty a_n$ do not tend to zero.

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