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Root Test

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Why did you need to learn about nth roots and algebra when you were in algebra class? It was so you could figure out when series converge, of course!

If you need to know if a series converges, but there is a power of \( n \) in it, then the Root Test is generally the go-to test. It can tell you if a series is absolutely convergent or divergent. This is different from most tests which tell you whether a series converges or diverges, but doesn't say anything about absolutely convergence.

One of the limits you will frequently need to apply the Root Test is

\[ \lim\limits_{n \to \infty} \frac{1}{\sqrt[n]{n}} = 1,\]

but why is that true. Showing that limit is actually equal to 1 uses the fact from properties of exponential functions and natural logs that

\[ e^{-\frac{\ln n}{n}} = \frac{1}{\sqrt[n]{n}}.\]

Since the exponential function is continuous,

\[ \begin{align} \lim\limits_{n \to \infty} e^{-\frac{\ln n}{n}} &= e^{-\lim\limits_{n \to \infty} \frac{\ln n}{n}} \\ &= e^{0} \\ &= 1, \end{align} \]

which gives you the desired result.

First, let's state the Root Test.

**Root Test:** Let

\[ \sum\limits_{n=1}^{\infty} a_n \]

be a series and define \( L \) by

\[ L = \lim\limits_{n \to \infty} \left| a_n \right|^{\frac{1}{n}}= \lim\limits_{n \to \infty} \sqrt[n]{\left| a_n \right|} .\]

Then the following hold:

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

2. If \( L > 1 \) then the series diverges.

3. If \( L = 1 \) then the test is inconclusive.

Notice that, unlike many series tests, there is no requirement that the terms of the series be positive. However, it can be challenging to apply the Root Test unless there is a power of \( n \) in the terms of the series. In the next section, you will see that the Root Test is also not very helpful if the series is conditionally convergent.

Remember that if a series converges absolutely, then it is, in fact, convergent. So if the Root Test tells you that a series converges absolutely, then it also tells you that it converges. Unfortunately, it will not tell you if a conditionally convergent series actually converges.

In fact the Root Test often can't be used on conditionally convergent series. Take for example the conditionally convergent alternating harmonic series

\[ \sum\limits_{n \to \infty} \frac{(-1)^n}{n} .\]

If you try to apply the Root Test, you get

\[ \begin{align} L &= \lim\limits_{n \to \infty} \left| a_n \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left| \frac{(-1)^n}{n} \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left( \frac{1}{n} \right)^{\frac{1}{n}} \\ &= 1. \end{align} \]

So in fact the Root Test doesn't tell you anything about the series. Instead to tell that the alternating harmonic series converges you would need to use the Alternating Series Test. For more details on that test, see Alternating Series.

The most significant rule about the Root Test is that it doesn't tell you anything if \( L = 1 \). In the previous section, you saw an example of a series that converges conditionally, but the Root Test couldn't tell you that because \( L = 1 \). Next, let's look at two more examples where the Root Test isn't helpful because \( L = 1 \).

If possible, use the Root Test to determine the convergence or divergence of the series

\[ \sum\limits_{n=1}^{\infty} \frac{1}{n^2}. \]

Answer:

This is a P-series with \( p = 2 \), so you already know it converges, and in fact it converges absolutely. But let's see what the Root Test gives you. If you take the limit,

\[ \begin{align} L &= \lim\limits_{n \to \infty} \left| a_n \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left| \frac{1}{n^2} \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left( \frac{1}{n^2} \right)^{\frac{1}{n}} \\ &= 1. \end{align} \]

So in fact the Root Test is inconclusive with this series.

If possible, use the Root Test to determine the convergence or divergence of the series

\[ \sum\limits_{n=1}^{\infty} \frac{1}{n^2}. \]

Answer:

This is a P-series with \( p = 1 \), or in other words the harmonic series, so you already know it diverges. If you take the limit to try and apply the Root Test,

\[ \begin{align} L &= \lim\limits_{n \to \infty} \left| a_n \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left| \frac{1}{n} \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left( \frac{1}{n} \right)^{\frac{1}{n}} \\ &= 1. \end{align} \]

So in fact the Root Test is inconclusive with this series.

Let's look at a couple of examples where the Root Test is useful.

If possible, determine the convergence or divergence of the series

\[ \sum\limits_{n=1}^{\infty} \frac{5^n}{n^n}. \]

Answer:

You might be tempted to use the Ratio Test for this problem instead of the Root Test. But the \( n^n \) in the denominator makes the Root Test a much better first attempt for looking at this series. Taking the limit,

\[ \begin{align} L &= \lim\limits_{n \to \infty} \left| a_n \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left| \frac{5^n}{n^n} \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left( \frac{5^n}{n^n} \right)^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \frac{5}{n} \\ &= 0 . \end{align} \]

Since \( L <1 \), the Root Test tells you that this series is absolutely convergent.

If possible, determine the convergence or divergence of the series

\[ \sum\limits_{n=1}^{\infty} \frac{(-6)^n}{n}. \]

Answer:

Given the power of \( n\) the Root Test is a good test to try for this series. Finding \( L \) gives:

\[ \begin{align} L &= \lim\limits_{n \to \infty} \left| a_n \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left| \frac{(-6)^n}{n} \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left( \frac{6^n}{n} \right)^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \frac{6}{n^{\frac{1}{n}}} \\ &= 6 . \end{align} \]

Since \( L > 1 \) the Root Test tells you that this series is divergent.

- \[ \lim\limits_{n \to \infty} \frac{1}{\sqrt[n]{n}} = 1\]
**Root Test:**Let\[ \sum\limits_{n=1}^{\infty} a_n \]

be a series and define \( L \) by

\[ L = \lim\limits_{n \to \infty} \left| a_n \right|^{\frac{1}{n}}= \lim\limits_{n \to \infty} \sqrt[n]{\left| a_n \right|} .\]

Then the following hold:

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

2. If \( L > 1 \) then the series diverges.

3. If \( L = 1 \) then the test is inconclusive.

The Root Test is used to tell if a series is absolutely convergent or divergent.

You don't solve a root test. It is a test to see if a series is absolutely convergent or divergent.

When the limit equals 1, the Root Test is inconclusive.

More about Root Test

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