Nth Root Test

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

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

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

Poof

 * 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 $$\sum_{n=1}^\infty |a_n|$$ 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 $$\sum_{n=1}^\infty a_n$$ do not tend to zero.

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