(ST7) Root Test#

In this lesson we are going to see how to:

  • use the Root Test to show that a series is convergent or divergent.

Review Video#

The Test#

Root Test

Given the series \(\displaystyle \sum_{n=1}^{\infty} a_n\), compute the limit: \(\displaystyle \lim_{n\to\infty}\sqrt[n]{\left|a_n\right|}=L\).

  • If \(L<1\), then the series is convergent (absolutely convergent).

  • If \(L>1\) or is infinite, then the series is divergent.

  • If \(L=1\), then this test is inconclusive.

Example 1#

Determine if the following series is convergent or divergent.

\[ \sum_{n=1}^{\infty} \dfrac{n^{3n}}{(2n+1)^{3n}} \]

(Click to see the steps.)

We start by rewriting our series so that everything is being raised to the power \(n\).

\[ \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \dfrac{n^{3n}}{(2n+1)^{3n}} = \sum_{n=1}^{\infty} \left(\dfrac{n}{2n+1}\right)^{3n} \]

The next step is to set up and simplify \(\sqrt[n]{|a_n|}\)

\[ \sqrt[n]{|a_n|} = \left|\left(\dfrac{n}{2n+1}\right)^{3n}\right|^{1/n} \]

Where we start by rewriting the root sign as a power function. Then since everything is positive, we can drop the absolute value sign and simplify the multiple exponents.

\[\begin{split} &= \left(\dfrac{n}{2n+1}\right)^{3n\cdot \tfrac{1}{n}} \\[10pt] &= \left(\dfrac{n}{2n+1}\right)^{3} \\[10pt] \end{split}\]

Now we are ready to actually calculate the limit:

\[ \lim_{n\to\infty}\sqrt[n]{|a_n|} =\lim_{n\to\infty}\left(\dfrac{n}{2n+1}\right)^{3} \]

And since the power function \((\cdot)^3\) is continuous, we can bring the limit inside the parentheses:

\[ =\left(\lim_{n\to\infty}\dfrac{n}{2n+1}\right)^3 \]

Do our usual technique of dividing every term by the highest power of \(n\) in the denominator:

\[\begin{split} =\left(\lim_{n\to\infty}\dfrac{\tfrac{n}{n}}{\tfrac{2n}{n}+\tfrac{1}{n}}\right)^3 \\[10pt] =\left(\lim_{n\to\infty}\dfrac{1}{2+\tfrac{1}{n}}\right)^3 \\[10pt] \end{split}\]

And calculate:

\[\begin{split} &=\left( \dfrac{1}{2+0}\right)^3 \\[10pt] &=\dfrac{1}{8}\\[10pt] \end{split}\]

Since the limit here is \(L=\tfrac{1}{8}\) which is less than \(1\), we can conclude by the root test that the series is convergent.

Inconclusive#

Note

The Root Test works best when you can write the series in the form: \(\sum a_n=\sum (b_n)^n\).

Remember that the Ratio Test and Root Test are both inconclusive when \(L=1\). This means the series might converge or it might diverge, we just don’t know; the test is inconclusive.

If the Root Test is inconclusive for a series, then the Ratio Test would also be inconclusive for that same series. (And the opposite is also true.)