# Root Test – Definition, Conditions, and Examples on Series

The root test is another helpful convergence test that is essential in our understanding of series convergence and divergence.  Learning about this test will add one more tool kit to our series and numerical analysis fundamentals.

The root test uses the $\boldsymbol{n}$th root of the $\boldsymbol{n}$th term of the series. We can determine the divergence or convergence of certain series by taking evaluating the limit of $\boldsymbol{\sqrt[n]{a_n}}$ as $\boldsymbol{n}$ approaches infinity.

Our discussion will cover the basics of the root test as well as the process of applying this convergence test. The root test is closely related to the ratio test, so refresh your understanding of the ratio test to better understand the root test.

## What is the root test?

The root test utilizes the $\boldsymbol{n}$th term of the series to determine the series’ divergence or convergence. The process of the root test is similar to the ratio test and this is why we say that the two tests are closely related. Instead of ratios, this time, we’re working with the $n$th root of the $n$th term.

\begin{aligned} \sum_{n = 1}^{\infty} a_n &= a_1 + a_2 + a_3 + …+ a_n +…\end{aligned}

We can use the $n$th root of $n$th term to determine the convergence or divergence of an infinite series such as the infinite series, $\sum_{n = 1}^{\infty} a_n$. The limit of the $n$th root of $a_n$ as $n \rightarrow \infty$ will determine whether the series is convergent or not.

If we let $L$ be the limit of $\sqrt[n]{a_n}$ as $n \rightarrow \infty$, the value of $L$ will determine the series is divergent or convergent.

### Root test rules

The root test rules look similar to that of the ratio test – we will check whether the resulting limit is less than, greater than, or equal to $1$.

 ROOT TEST RULESSuppose that $\lim_{n \rightarrow \infty} \sqrt[n]{|a_n|} = L$, the root test rule states that:i)             If $\boldsymbol{L < 1}$, the series $\boldsymbol{\sum_{n =1}^{\infty} a_n}$ is absolutely convergent (and consequently, convergent).ii)            If $\boldsymbol{L > 1}$, the series $\boldsymbol{\sum_{n =1}^{\infty} a_n}$ is divergent.iii)           If $\boldsymbol{L = 1}$, the series $\boldsymbol{\sum_{n =1}^{\infty} a_n}$ may be divergent, conditionally convergent, and absolutely divergent.

## When to use the root test?

The best time to apply the root test is when we’re working with series that contains terms involving exponents. Since we’re taking the $n$th root or raising the $n$th term to a power of $\dfrac{1}{n}$, series with expressions raised to $n$ will benefit the most from the root test.

\begin{aligned}&\sum_{n = 1}^{\infty} \dfrac{n^n}{4 ^{2n + 1}}\\&\sum_{n = 1}^{\infty} \left(\dfrac{2n + 1}{3n + 4} \right )^n\\&\sum_{n =1}^{\infty} (-1)^n \left(\dfrac{6}{n – 4} \right )^n\end{aligned}

These are examples of series where we can easily apply the root test to check the series for convergence or divergence.  A rule of thumb to remember: when the process gets easier because of the root test, then you’re using the right convergence test.

Now that we’ve covered the fundamentals of the root test, let’s now learn how to apply the root test on actual series.

How to do the root test?

When applying the root test, the first goal is to express the $n$th as a power of $n$ or $kn$, where $k$ is a real number. Now, use the pointers below to apply the root test properly.

• Find the expression for $\sqrt[n]{a_n}$ or $(a_n)^{\frac{1}{n}}$ then simplify the expression.
• Evaluate the limit of the $n$th root of $a_n$ as $n \rightarrow \infty$.
• Apply the appropriate limit laws and exponent rules to simplify the resulting expression for $\sqrt[n]{a_n}$.
• Now, check if $L = \lim_{n \rightarrow \infty} \sqrt[n]{a_n}$ is less than, greater than or equal to $1$.
 \begin{aligned}\boldsymbol{L <1}\end{aligned} \begin{aligned}\boldsymbol{L >1}\end{aligned} \begin{aligned}\boldsymbol{L =}\end{aligned} Series converges Series diverges Inconclusive

To help you better understand these steps, let’s go ahead and confirm that the series, $\sum_{n = 1}^{\infty} \dfrac{e^{2n}}{n^n}$, converges. We can rewrite $\dfrac{e^{2n}}{n^n}$ as $\left(\dfrac{e^2}{n}\right)^n$ then take its $n$th root as shown below.

\begin{aligned}\sqrt[n]{a_n} &= \sqrt[n]{\left(\dfrac{e^2}{n}\right)^n}\\&= \left[\left(\dfrac{e^2}{n}\right)^n \right ]^{1/n}\\&= \left(\dfrac{e^2}{n}\right)^{n \cdot (1/n)}\\&= \dfrac{e^2}{n} \end{aligned}

Now let’s evaluate the simplified expression of $\sqrt[n]{a_n}$’s limit as $n \rightarrow \infty$. We’ve learned in the past that $\lim_{n \rightarrow \infty}\dfrac{c}{n^k} = 0$, where both $c$ and $k$ are constants.

\begin{aligned} L &= \lim_{n \rightarrow \infty} \sqrt[n]{a_n} \\&= \lim_{n \rightarrow \infty} \dfrac{e^2}{n}\\&= 0\\&< 1 \end{aligned}

Since $L < 1$, this confirms that the series is convergent through the root test.

Are you ready to work on more problems? We’ve prepared more for you to master this convergence test. Head over to the next section when you’re ready!

Example 1

Determine whether the series, $\sum_{n = 1}^{\infty} \left(\dfrac{3n – 1}{4n + 2}\right)^n$, is convergent or divergent.

Solution

Since the $n$th term, $a_n = \left(\dfrac{3n – 1}{4n + 2}\right)^n$,  contains an exponent of $n$, we can apply the root test to check this series for its convergence or divergence. Let’s begin by taking the $n$th root of $a_n$ as shown below.

\begin{aligned}\sqrt[n]{a_n} &= \sqrt[n]{\left(\dfrac{3n – 1}{4n + 2}\right)^n}\\&= \left[\left(\dfrac{3n – 1}{4n + 2}\right)^n \right ]^{1/n}\\&= \dfrac{3n – 1}{4n + 2} \end{aligned}

Find the limit of $\dfrac{3n – 1}{4n + 2}$ as $n \rightarrow \infty$ to determine whether the series is divergent or convergent.

\begin{aligned} L &= \lim_{n \rightarrow \infty} \sqrt[n]{a_n} \\&= \lim_{n \rightarrow \infty} \dfrac{3n – 1}{4n + 2}\\&=\lim_{n \rightarrow \infty} \dfrac{\dfrac{3n}{n} – \dfrac{1}{n}}{\dfrac{4n}{n} + \dfrac{2}{n}}\\&= \lim_{n \rightarrow \infty} \dfrac{3 – \dfrac{1}{n}}{4 + \dfrac{2}{n}}\\&= \dfrac{3 -0}{4 + 0}\\&= \dfrac{3}{4}\\&< 1 \end{aligned}

Since $L < 1$, we can confirm that $\sum_{n = 1}^{\infty} \left(\dfrac{3n – 1}{4n + 2}\right)^n$ is convergent.

Example 2

Determine whether the series, $\sum_{n = 1}^{\infty} \dfrac{n^n}{(\ln n)^n}$, is convergent or divergent.

Solution

Rewrite $\dfrac{n^n}{(\ln n)^n}$ as $\left(\dfrac{n}{\ln n}\right)^n$ then take the $n$th root of this expression.

\begin{aligned}\sqrt[n]{a_n} &= \sqrt[n]{\left(\dfrac{n}{\ln n}\right)^n}\\&= \left[\left(\dfrac{n}{\ln n}\right)^n\right ]^{1/n}\\&= \dfrac{n}{\ln n}\end{aligned}

If we evaluate the limit of $\sqrt[n]{a_n}$ as $n \rightarrow \infty$, the initial result will be $\dfrac{\infty}{\infty}$. Apply the L’Hopitals rule by taking the numerator and denominator’s derivatives then evaluating the limit of the resulting expression instead.

\begin{aligned} L &= \lim_{n \rightarrow \infty} \sqrt[n]{a_n} \\&= \lim_{n \rightarrow \infty} \dfrac{n}{\ln n}\\&=\lim_{n \rightarrow \infty} \dfrac{\dfrac{d}{dn} (n)}{\dfrac{d}{dn}(\ln n)}\\&= \lim_{n \rightarrow \infty} \dfrac{1}{\dfrac{1}{n}}\\&= \lim_{n \rightarrow \infty} n\\&= \infty\\ &> 1 \end{aligned}

From this, we can see that $L >1$, so through the root test, we can confirm that the series, $\sum_{n = 1}^{\infty} \dfrac{n^n}{(\ln n)^n}$, is divergent.

### Practice Questions

1. Determine whether the series, $\sum_{n = 1}^{\infty} \left(\dfrac{4n +5}{6n – 1}\right)^n$, is convergent or divergent.

2. Determine whether the series, $\sum_{n = 1}^{\infty} \dfrac{n^n}{3^n}$, is convergent or divergent.
3. Determine whether the series, $\sum_{n = 1}^{\infty} \left(\dfrac{4n +5}{6n – 1}\right)^n$, is convergent or divergent.

4. Determine whether the series, $\sum_{n = 1}^{\infty} \left(\dfrac{1}{n + 1} + \dfrac{1}{n + 2} + \dfrac{1}{n + 3}+… +\dfrac{1}{3n} \right)^n$, is convergent or divergent.

1. Since $L = \dfrac{2}{3} <1$, $\sum_{n = 1}^{\infty} \left(\dfrac{4n +5}{6n – 1}\right)^n$ is convergent.
2. Since $L = \infty >1$, $\sum_{n = 1}^{\infty} \dfrac{n^n}{3^n}$ is divergent.
3. Since $L = 0 <1$, $\sum_{n = 1}^{\infty} \dfrac{n^n}{3^n}$ is convergent.
4. Since $L = \ln 3 >1$, $\sum_{n = 1}^{\infty} \left(\dfrac{1}{n + 1} + \dfrac{1}{n + 2} + \dfrac{1}{n + 3}+… +\dfrac{1}{3n} \right)^n$ is convergent.