Term test – Serlo

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In this chapter we will see an easy and useful criterion for divergence. This criterion is called term test or also null-sequence test or divergence test. It says that every series , where is not a null sequence, is divergent. When you invert that, this means that for every convergent series we have that( is a null sequence).

Term test[Bearbeiten]

Theorem (Term test)

If a series is convergent, then is a null sequence. That means, that every series is divergent, if is divergent or .

Example (Term test)

The series is divergent, because is divergent ( and are both accumulation points and thus there is no limit).

Also the series diverges, because .

Warning

The criterion that is a null sequence is only a necessary but no sufficient criterion for the convergences of the series .

This means: from the fact that we cannot follow that converges. For instance the harmonic series is divergent although .

Proof with telescoping sums[Bearbeiten]

Proof (Term test)

Our premise is that is a convergent series. We want to conclude that is a null sequence.

A member of the sequence can be written as the difference between two consecutive partial sums and :

From our premise we know that has a limit . Thus we have

But also , because the limit will not change if we simply shift indexes. Put together we obtain:

We can conclude that must be null sequence.

Proof using Cauchy criterion[Bearbeiten]

Proof (Term test)

We can proof the same result using the Cauchy criterion. As a reminder, every convergent series satisfies the Cauchy criterion:

We don't consider all , but only the case :

The last formula is exactly the -definition for what it means that is a null sequences. In other words we have showed that .

Example[Bearbeiten]

Beispielaufgabe zum Trivialkriterium

Exercise

Show that is divergent.

Solution

We have

From the above we see that is not a null sequence. So the series is divergent according to the term test.

Outlook: Stronger version of term test[Bearbeiten]

If we demand that is monotonically decreasing, then we can show that even is a null sequence. See the respective exercise.