Root test – "Math for Non-Geeks"

verweis=Math for Non-Geeks

Let us turn our attention to the root test, which is a powerful tool for proving the convergence or divergence of a given series. It is based off the direct comparison test, in fact we will compare the series with the geometric series $\sum_{k = 1}^{\infty} q^{k}$ with $0 \leq q < 1$ .

The root test was first published in 1821 in the textbook „Cours d'analyse“ from French mathematician Augustin Louis Cauchy ^[1].

Derivation

Recap

We have already learned how to use the direct comparison test with a majorant. In summary, a series $\sum_{k = 1}^{\infty} a_{k}$ with $a_{k} \geq 0$ is convergent, if there exists a convergent series $\sum_{k = 1}^{\infty} b_{k}$ with $a_{k} \leq b_{k}$ .

Furthermore, we know that every geometric series $\sum_{k = 1}^{\infty} q^{k}$ with $q \in [0, 1)$ is convergent.

First derivation

Let $\sum_{k = 1}^{\infty} a_{k}$ be a series. Also let $a_{k} \geq 0$ for all $k \in ℕ$ , because this is a constraint we need for the direct comparison test. For our majorant we need a $q \in [0, 1)$ with $a_{k} \leq q^{k}$ . Then we have

$\begin{matrix} \sum_{k = 1}^{\infty} a_{k} & \leq \sum_{k = 1}^{\infty} q^{k} = \sum_{k = 0}^{\infty} q^{k} - q^{0} \\ = \frac{1}{1 - q} - 1 = \frac{q}{1 - q} < \infty \end{matrix}$

The series $\sum_{k = 1}^{\infty} a_{k}$ is convergent by our direct comparison test. The inequality $a_{k} \leq q^{k}$ can be transformed:

$a_{k} \leq q^{k} ⟺ \sqrt[k]{a_{k}} \leq q$

Thus, if there exists a $q$ with $0 \leq q < 1$ , so that $\sqrt[k]{a_{k}} \leq q$ , then $a_{k} \leq q^{k}$ and the series $\sum_{k = 1}^{\infty} a_{k}$ is convergent.

Comprehension question: We start the geometric series

\sum_{k = 1}^{\infty} q^{k}

at

k = 1

and not (as usual) at

k = 0

. Why does this make sense in the above derivation?

In the step $a_{k} \leq q^{k} ⟺ \sqrt[k]{a_{k}} \leq q$ we take the $k$ -th root, but the $0$ -th root is not defined, because that would mean $q^{\frac{1}{0}}$ . This is why $k$ should start at $1$ .

Bringing limit superior into play

For the convergence behaviour we can ignore the value of finitely many summands. Thus $\sqrt[k]{a_{k}} \leq q$ must not be satisfied for all $k \in ℕ$ , but for all $k \in ℕ$ , with finitely many exceptions. So the inequality $\sqrt[k]{a_{k}} \leq q$ must be satisfied for almost all $k \in ℕ$ .

We can restate the requirement, that there must exist a $q \in [0, 1)$ with $\sqrt[k]{a_{k}} \leq q$ for almost all $k \in ℕ$ , using the limit superior:

$\exists \tilde{q} \in [0, 1) : \underset{k \to \infty}{lim sup} \sqrt[k]{a_{k}} \leq \tilde{q}$

Or in other terms:

$\underset{k \to \infty}{lim sup} \sqrt[k]{a_{k}} < 1$

If $\sqrt[k]{a_{k}} \leq q$ for almost all $k$ , the succession ${(\sqrt[k]{a_{k}})}_{k \in ℕ}$ is bounded above and must have an accumulation point less than or equal to $q$ . This accumulation point is equal to $\underset{k \to \infty}{lim sup} \sqrt[k]{a_{k}}$ and $\underset{k \to \infty}{lim sup} \sqrt[k]{a_{k}} \leq q$ .

Conversely, let $\underset{k \to \infty}{lim sup} \sqrt[k]{a_{k}} \leq q$ for a $q \in [0, 1)$ . Then for all $ϵ > 0$ the inequality $\sqrt[k]{a_{k}} \leq q + ϵ$ is satisfied for almost all $k \in ℕ$ . Because $q < 1$ there exists $ϵ > 0$ , that is small enough so that $q + ϵ < 1$ . Set $\tilde{q} = q + ϵ$ . We have $\tilde{q} < 1$ and the inequality $\sqrt[k]{a_{k}} \leq \tilde{q}$ holds for almost all $k$ .

Summary: Instead of $\sqrt[k]{a_{k}} \leq q$ for almost all $k \in ℕ$ it suffices to show $\underset{k \to \infty}{lim sup} \sqrt[k]{a_{k}} < 1$ , to prove convergence the series.

What about absolute convergence?

What happens if not all $a_{k} \geq 0$ ? Then we cannot use the above argumentation, because for even $k$ and $a_{k} < 0$ the values $\sqrt[k]{a_{k}}$ is not defined. But we still have a valid argumentation if we talk about $\sum_{k = 1}^{\infty} | a_{k} |$ , to show the absolute convergence of the series (from which normal convergence follows immediately). Thus, for series with $a_{k} \geq 0$ for all $k$ we have $| a_{k} | = a_{k}$ . In this case nothing changes because for $a_{k}$ we can also use $| a_{k} |$ . The conclusion is:

If

\underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} < 1

, then the series

\sum_{k = 1}^{\infty} a_{k}

converges absolutely.

Root test for divergence

We have derived the root test to show the convergence of a series. Can we also derive a criterion for divergence? Let us assume that $\underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} > 1$ . Then for infinitely many $k \in ℕ$ we have the inequality $\sqrt[k]{| a_{k} |} \geq 1$ . For these $k$ we have $| a_{k} | \geq 1^{k} = 1$ , thus ${(| a_{k} |)}_{k \in ℕ}$ cannot be a null sequence. But then ${(a_{k})}_{k \in ℕ}$ is also no null sequence. From the term test, it follows immediately that $\sum_{k = 1}^{\infty} a_{k}$ is divergent. We can generalize this case if instead of $\underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} > 1$ we require the inequality $\sqrt[k]{| a_{k} |} \geq 1$ for almost all $k \in ℕ$ .

Theorem

Theorem (Root test)

Let $\sum_{k = 1}^{\infty} a_{k}$ be a series. If $\underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} < 1$ , then the series is absolutely convergent. If $\underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} > 1$ , then it is divergent. Even if $\sqrt[k]{| a_{k} |} \geq 1$ for infinitely many $k \in ℕ$ , the series diverges.

The proof uses the same ideas as the derivation above:

Proof (Root test)

Proof step: From $\underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} < 1$ follows the absolute convergence of $\sum_{k = 1}^{\infty} a_{k}$ .

Let $\sum_{k = 1}^{\infty} a_{k}$ be a series. If $\underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} < 1$ , then $\underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} \leq q$ for a $0 \leq q < 1$ (we can choose $q = \underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |}$ for example).

Choose $ϵ > 0$ small enough, so that $q + ϵ < 1$ . From the definition of limit superior it follows that for almost all $k \in ℕ$ the inequality $\sqrt[k]{| a_{k} |} \leq q + ϵ$ holds. But that means $| a_{k} | \leq (q + ϵ)^{k}$ for almost all $k$ . Since $\sum_{k = 1}^{\infty} (q + ϵ)^{k}$ is convergent (it is a geometric series with $q + ϵ < 1$ ) also the series $\sum_{k = 1}^{\infty} | a_{k} |$ is convergent by direct comparison. So the series $\sum_{k = 1}^{\infty} a_{k}$ converges absolutely.

Proof step: From $\underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} > 1$ or $\sqrt[k]{| a_{k} |} \geq 1$ for infinitely many $k$ follows the divergence of $\sum_{k = 1}^{\infty} a_{k}$ .

Let $\underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} > 1$ or $\sqrt[k]{| a_{k} |} \geq 1$ for infinitely many $k$ . This implies $| a_{k} | \geq 1^{k} = 1$ for infinitely many $k$ . But then ${(| a_{k} |)}_{k \in ℕ}$ cannot be a null sequence and thus ${(a_{k})}_{k \in ℕ}$ is also not a null sequence. From the term test we know that $\sum_{k = 1}^{\infty} a_{k}$ is divergent.

Hint

If ${(\sqrt[k]{| a_{k} |})}_{k \in ℕ}$ is convergent, then $lim_{k \to \infty} \sqrt[k]{| a_{k} |} = \underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |}$ . Therefore, we could also consider the limit $lim_{k \to \infty} \sqrt[k]{| a_{k} |}$ if it exists. This is usually done when proving convergence using the root test.

Limitations of the root test

In case $\underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} = 1$ we cannot say whether we have convergence or divergence. In fact there exist both convergent and divergent series that satisfy this equation. First consider the harmonic series $\sum_{k = 1}^{\infty} \frac{1}{k}$ , which is divergent. We have

$\begin{matrix} \underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} & = \underset{k \to \infty}{lim sup} \sqrt[k]{| \frac{1}{k} |} \\ = \underset{k \to \infty}{lim sup} \sqrt[k]{\frac{1}{k}} \\ = \underset{k \to \infty}{lim sup} \frac{1}{\sqrt[k]{k}} \\ ↓ lim_{k \to \infty} \sqrt[k]{k} = 1 \\ = \frac{1}{1} = 1 \end{matrix}$

But also the convergent series $\sum_{k = 1}^{\infty} \frac{1}{k^{2}}$ satisfies this equation:

$\begin{matrix} \underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} & = \underset{k \to \infty}{lim sup} \sqrt[k]{| \frac{1}{k^{2}} |} \\ = \underset{k \to \infty}{lim sup} \sqrt[k]{\frac{1}{k^{2}}} \\ = \underset{k \to \infty}{lim sup} \frac{1}{\sqrt[k]{k^{2}}} \\ = \underset{k \to \infty}{lim sup} {(\frac{1}{\sqrt[k]{k}})}^{2} \\ ↓ lim_{k \to \infty} \sqrt[k]{k} = 1 \\ = {(\frac{1}{1})}^{2} = 1 \end{matrix}$

These examples show that we cannot conclude convergence or divergence from $\underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} = 1$ . In this case we have to employ another convergence criterion!

How to apply the root test

To apply the root test on a series $\sum_{k = 1}^{\infty} a_{k}$ we can proceed as follows: We compute $\sqrt[k]{| a_{k} |}$ and find the limit (if the limit exists) or the limit superior.

If $\underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} < 1$ , then the series converges absolutely.
If $\underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} > 1$ , then the series diverges.
If $\sqrt[k]{| a_{k} |} \geq 1$ for infinitely many $k$ , then the series diverges.
Else, if none of the above is true, we cannot make a statement about convergence behaviour using the root test.

Exercises

Exercise 1

Exercise

Is the series $\sum_{k = 1}^{\infty} \frac{k^{3}}{3^{k}}$ convergent or divergent?

Solution

We compute the limit of $\sqrt[k]{| \frac{k^{3}}{3^{k}} |}$ :

$\begin{matrix} \underset{k \to \infty}{lim sup} \sqrt[k]{| \frac{k^{3}}{3^{k}} |} & = \underset{k \to \infty}{lim sup} \sqrt[k]{\frac{k^{3}}{3^{k}}} \\ = \underset{k \to \infty}{lim sup} \frac{\sqrt[k]{k^{3}}}{\sqrt[k]{3^{k}}} \\ = \underset{k \to \infty}{lim sup} \frac{{\sqrt[k]{k}}^{3}}{3} \\ ↓ lim_{k \to \infty} \sqrt[k]{k} = 1 \\ = \frac{1^{3}}{3} = \frac{1}{3} < 1 \end{matrix}$

Because $\underset{k \to \infty}{lim sup} \sqrt[k]{| \frac{k^{3}}{3^{k}} |} < 1$ it follows from the root test that the series is convergent.

Exercise 2

Exercise

Is the series $\sum_{k = 1}^{\infty} {(\frac{4 k + 5}{2 k + 3})}^{k}$ convergent or divergent?

Solution

We have

$\begin{matrix} \underset{k \to \infty}{lim sup} \sqrt[k]{{(\frac{4 k + 5}{2 k + 3})}^{k}} & = \underset{k \to \infty}{lim sup} \frac{4 k + 5}{2 k + 3} \\ = \underset{k \to \infty}{lim sup} \frac{4 + \frac{5}{k}}{2 + \frac{3}{k}} \\ = \frac{4}{2} = 2 > 1 \end{matrix}$

Because $\underset{k \to \infty}{lim sup} \sqrt[k]{{(\frac{4 k + 5}{2 k + 3})}^{k}} > 1$ the series is divergent according to the root test.

Einzelnachweise

↑ Siehe the answer to the question „Where is the root test first proved“ of Q&A website „History of Science and Mathematics“

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[1] Siehe the answer to the question „Where is the root test first proved“ of Q&A website „History of Science and Mathematics“

[1]