Nth Root Test

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Theorem

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

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

Then:

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


Proof

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 $\langle {b_n} \rangle = \langle \left \vert {a_n} \right \vert \rangle$.

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

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

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

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

$\Box$


Divergence

Let $l > 1$.

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

Seeking a contradiction, suppose that there exist an upper bound for the set:

$S := \left\{{n \in \N: \left\vert {a_n} \right\vert^{1/n} > l - \epsilon}\right\}$

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 $l$.

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

This means that there exist arbitrarily large $n$ such that:

$\left\vert {a_n} \right\vert > \left( l - \epsilon \right)^n$

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

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

$\blacksquare$


Notes

If $l = 1$, it is impossible to say, without further analysis, whether $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges absolutely, converges conditionally, or diverges.

If $\displaystyle \limsup_{n \to \infty} \left\vert {a_n}\right\vert^{1/n} = \infty$, then of course $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ diverges.


Sources