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

 
Decision tree for the root test

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.