Root test – Serlo

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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 with .

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

Derivation[Bearbeiten]

Recap[Bearbeiten]

We have already learned how to use the direct comparison test with a majorant. In summary, a series with is convergent, if there exists a convergent series with .

Furthermore we know that every geometric series with is convergent.

First derivation[Bearbeiten]

Let be a series. Also let for all , because this is a constraint we need for the direct comparison test. For our majorant we need a with . Then we have

The series is convergent by our direct comparison test. The inequality can be transformed:

Thus if there exists a with , so that , then and the series is convergent.

Comprehension question: We start the geometric series at and not (as usual) at . Why does this make sense in the above derivation?

In the step we take the -th root, but the -th root is not defined, because that would mean . This is why should start at .

Bringing limit superior into play[Bearbeiten]

For the convergence behaviour we can ignore the value of finitely many summands. Thus must not be satisfied for all , but for all , with finitely many exceptions. So the inequality must be satisfied for almost all .

We can restate the requirement, that there must exist a with for almost all , using the limit superior:

Or in other terms:

If for almost all , the succession is bounded above and must have an accumulation point less than or equal to . This accumulation point is equal to and .

Conversely, let for a . Then for all all the inequality is satisfied for almost all . Because there exists , that is small enough so that . Set . We have and the inequality holds for almost all .

Summary: Instead of for almost all it suffices to show , to prove convergence the series.

What about absolute convergence?[Bearbeiten]

What happens if not all ? Then we cannot use the above argumentation, because for even and the values is not defined. But we still have a valid argumentation if we talk about , to show the absolute convergence of the series (from which normal convergence follows immediately). Thus for series with for all we have . In this case nothing changes because for we can also use . The conclusion is:

If , then the series converges absolutely.

Root test for divergence[Bearbeiten]

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 . Then for infinitely many we have the inequality . For these we have , thus cannot be a null sequence. But then is also no null sequence. From the term test it follows immediately that is divergent. We can generalize this case if instead of we require the inequality for almost all .

Theorem[Bearbeiten]

Theorem (Root test)

Let be a series. If , then the series is absolutely convergent. If , then it is divergent. Even if for infinitely many , the series diverges.

The proof uses the same ideas as the derivation above:

Proof (Root test)

Proof step: From follows the absolute convergence of .

Let be a series. If , then for a (we can choose for example).

Choose small enough, so that . From the definition of limit superior it follows that for almost all the inequality holds. But that means for almost all . Since is convergent (it is a geometric series with ) also the series is convergent by direct comparison. So the series converges absolutely.

Proof step: From or for infinitely many follows the divergence of .

Let or for infinitely many . This implies for infinitely many . But then cannot be a null sequence and thus is also not a null sequence. From the term test we know that is divergent.

Hint

If is convergent, then . Therefore we could also consider the limit if it exists. This is usually done when proving convergence using the root test.

Limitations of the root test[Bearbeiten]

In case 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 , which is divergent. We have

But also the convergent series satisfies this equation:

These examples show that we cannot conclude convergence or divergence from . In this case we have to employ another convergence criterion!

How to apply the root test[Bearbeiten]

Entscheidungsbaum für das Wurzelkriterium

To apply the root test on a series we can proceed as follows: We compute and find the limit (if the limit exists) or the limit superior.

  1. If , then the series converges absolutely.
  2. If , then the series diverges.
  3. If for infinitely many , then the series diverges.
  4. Else, if none of the above is true, we cannot make a statement about convergence behaviour using the root test.

Exercises[Bearbeiten]

Exercise 1[Bearbeiten]

Exercise

Is the series convergent or divergent?

Solution

We compute the limit of :

Because it follows from the root test that the series is convergent.

Exercise 2[Bearbeiten]

Exercise

Is the series convergent or divergent?

Solution

We have

Because the series is divergent according to the root test.