The Ratio and Root Tests

$\displaystyle\sqrt[n]{\left|a_n\right|}=\left|a_n\right|^{1/n}$ , 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.

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The Ratio and Root Tests

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The Root Test

An important limit you may need in order to use the Root Test

Justification of the Root Test