The Root Test involves looking at
$\displaystyle\lim_{n\to\infty}\sqrt[n]{\left|a_n\right|}$, hence
the name.
Notice: $\displaystyle\sqrt[n]{\left|a_n\right|}=\left|a_n\right|^{1/n}$,
and you will see both notations.
The Root Test, like the Ratio Test, is a test to determine absolute
convergence (or not). While the Ratio Test is good to use with
factorials, since there is that lovely cancellation of terms of
factorials when you look at ratios, the
Root Test is best used when there are terms to the $n^{th}$
power with no factorials.
The Root Test: Suppose that
$\displaystyle\lim_{n \to
\infty}\sqrt[n]{\left|a_n\right|}=L$.
If $L<1$, then $\sum a_n$ converges absolutely.
If $L>1$, or the limit goes to $\infty$, then $\sum
a_n$ diverges.
If $L=1$, or $L$ does not exist, then the test is
inconclusive and we say the Ratio Test fails.
We must do more work to determine convergence,
Example: Test the absolute/conditional convergence of
the series
$\displaystyle\sum_{n=1}^\infty(-1)^n\left(\frac{2n-7}{5n+2}\right)^n$.
Solution:We run through our
tests: Is this a geometric series? No, we
have a function raised to the $n^{th}$ power, not a number (but we
get a glimmer of something here, right?). Our terms are
alternating, but the AST will not tell us whether or not we have
absolute convergence. We don't want to think about integrating
this expression. We could use the Ratio Test, but since our
terms are raised to the $n^{th}$ power, we decide to try the Root
Test. You will see how nicely it works with powers:
As $n\to\infty$,
$\displaystyle\sqrt[n]{\left|a_n\right|}=\sqrt[n]{\left(\frac{2n-7}{5n+2}\right)^n}=\frac{2n-7}{5n+2}=\frac{2-\frac7n}{5+\frac2n}\longrightarrow\frac25$.
Since $\frac25<1$, our series converges absolutely.
It is important to consider our litany of
convergence/divergence tests before doing work. It can
save valuable time, as well as helping you begin to recognize
which test to do when. On an exam, you will not know which
module the series came from!
An important limit you may need in order to use the Root Test
You may have to compute the limit of sequences like $\sqrt[n]n$
or $\lim\sqrt[n]{n^2}$. Let's
compute this now, so we can use it later. Let
$a$ be a positive real number. First notice that
$\displaystyle\lim_{n\to\infty}\sqrt[n]{n^a}=\lim_{n\to\infty}\left(n^a\right)^{1/n}=\lim_{n\to\infty}n^{a/n}$
looks like the indeterminate form $\infty^0$, so it can best be
computed by computing the limit of the logarithm of this
expression. We will use the function with continuous
variable $x$ so that we can take derivatives using l'Hospital's
Rule.
$\displaystyle\lim_{x\to\infty}\ln\left(x^{a/x}\right)=\lim_{x\to\infty}\frac
ax\cdot\ln x \underset{\,\\
0\cdot\infty}{=}\lim_{x\to\infty}\frac{\ln x}{\frac xa}\underset{
\,\\ \frac{\infty}{\infty},\text{
l'H}}{=}\lim_{x\to\infty}\frac{\frac 1x}{\frac
1a}=\lim_{x\to\infty}\frac ax=0$,
so $\displaystyle\lim_{x\to\infty}x^{a/x}=e^0=1$. We have
shown:
A Handy Limit
For any positive real number $a$,
$$\lim_{n\to\infty}\sqrt[n]{n^a}=\lim_{n\to\infty}n^{a/n}=1.$$
Justification of the Root Test
Notice that we can think of the relationship of
$\sqrt[n]{\left|a_n\right|}$ to the terms of a geometric series,
since $\sqrt[n]{r^n} = r$. The video gives a justification,
using this idea, of why the Root Test works.