Root Test
The root test uses a similar approach to the ratio test, but focuses on the [latex]n[/latex]th root of terms rather than their ratios.
Consider a series [latex]\displaystyle\sum_{n=1}^{\infty}{a}{n}[/latex] where [latex]\underset{n\to \infty }{\text{lim}}\sqrt[n]{|a_{n}|}=\rho[/latex] for some real number [latex]\rho[/latex]. For sufficiently large [latex]N[/latex], we have [latex]|a_{N}|\approx \rho^{N}[/latex].
This means we can approximate the tail of our series:
[latex]|a_{N}|+|a_{N+1}|+|a_{N+2}|+\cdots \approx \rho^{N}+\rho^{N+1}+\rho^{N+2}+\cdots[/latex]
The right side is a geometric series, which gives us the same convergence pattern as the ratio test.
root test
Consider the series [latex]\displaystyle\sum_{n=1}^{\infty}{a}_{n}[/latex]. Let
[latex]\rho =\underset{n\to \infty }{\text{lim}}\sqrt[n]{|a_{n}|}[/latex]
Then:
- If [latex]0\leq \rho <1[/latex], the series [latex]\displaystyle\sum_{n=1}^{\infty}{a}_{n}[/latex] converges absolutely
- If [latex]\rho >1[/latex] or [latex]\rho =\infty[/latex], the series [latex]\displaystyle\sum_{n=1}^{\infty}{a}_{n}[/latex] diverges
- If [latex]\rho =1[/latex], the test is inconclusive and does not provide any information
Why the Root Test Sometimes Fails
For any p-series [latex]\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^p}[/latex], we get [latex]\rho = 1[/latex] regardless of the value of [latex]p[/latex]. Yet we know p-series converge when [latex]p > 1[/latex] and diverge when [latex]p \leq 1[/latex]. This shows why [latex]\rho = 1[/latex] makes the test inconclusive.
When to Use the Root Test
The root test works particularly well for series whose terms involve exponentials, especially when [latex]|a_{n}|=b_{n}^{n}[/latex] for some sequence [latex]{b_{n}}[/latex]. In these cases, [latex]\sqrt[n]{|a_{n}|}=b_{n}[/latex], so you only need to evaluate [latex]\underset{n\to \infty }{\text{lim}}b_{n}[/latex].
For each of the following series, use the root test to determine whether the series converges or diverges.
- [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\left({n}^{2}+3n\right)}^{n}}{{\left(4{n}^{2}+5\right)}^{n}}[/latex]
- [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{n}^{n}}{{\left(\text{ln}\left(n\right)\right)}^{n}}[/latex]