# Bounded series and convergence – Serlo

## Bounded series with positive summands converge[Bearbeiten]

As we learned in the chapter „Monotoniekriterium für Folgen“,^{[1]} monotonicity and boundedness combined are a sufficient criterion to show convergence of a sequence.
This criterion is also applicable to series, since the meaning of a series is defined by the *sequence* of its partial sums.
In the following, we will derive a convergence theorem for series that is based on the monotonicity and boundedness criterion for sequences.

We proceed in two steps by separately examining the monotonicity and boundedness of partial sums of some series .

First we ask: Under what circumstances is the sequence of partial sums monotonically increasing? Starting with an arbitrary partial sum, we obtain the next one by adding the corresponding term of the series. Therefore the next partial sum is greater (or equal), whenever this summand is positive (or zero). If we require monotonicity for the partial sums, this means all terms of the series must be non-negative (except for the first one, which does not play the role of a difference between partial sums).

We now want to show this statement formally. The relation between two subsequent partial sums is

The sequence is monotonically increasing, if for all . Using the relation we just derived, we find:

Hence the sequence of partial sums is monotonically increasing, if and only if for all . This simply means for all .

Next, we address boundedness of the sequence . A real sequence is bounded, if there exists some constant , such that for all . With the being partial sums, we can picture this in the following way: Each partial sum is a truncation of the infinite sum , where we drop all terms after . Each partial sum is just a finite sum of finite terms, so it is clearly finite itself. Therefore it is not hard to find a common upper bound for some of the partial sums. However, to show boundedness of the sequence, we need to find an upper bound for *all* partial sums at once. So we are looking for a number that is greater than any of the truncated sums, no matter how late we do the truncation.

We summarize these observations in the following theorem:

**Theorem** (Bounded series with non-negative summands converge)

Let be a series with for all . This series converges, if and only if there exists some bound , such that for all .

**Proof** (Bounded series with non-negative summands converge)

Let be a series with for all .

**Proof step:** If the sequence is bounded, the series converges.

We have shown above that for a series with non-negative summands the sequence of partial sums is monotonically increasing. According to the monotonicity criterion for sequences, a monotonically increasing sequence with an upper bound converges. By definition, if its sequence of partial sums converges, a series is said to converge.

**Proof step:** If converges, its partial sums are bounded.

By definition, if the series converges, its sequence of partial sums converges. In chapter „Unbeschränkte Folgen divergieren“^{[2]} we learned that convergent sequences are bounded.

Analogously, we can also proof a theorem for monotonically decreasing partial sums:

**Theorem** (Bounded series with non-positive summands converge)

Let be a series with for all . This series converges, if and only if there exists some bound , such that for all .

**Hint**

As we have seen, both theorems are applications of the monotonicity criterion for sequences. Therefore, the literature often refers to them as the monotonicity criterion for series.

## Application: Convergence of the hyperharmonic series [Bearbeiten]

**Theorem** (Convergence of the hyperharmonic series)

The hyperharmonic series is convergent for (for we obtain the harmonic series, whose divergence has been shown in the corresponding article^{[3]}).

**How to get to the proof?** (Convergence of the hyperharmonic series)

We want to proof convergence by applying the theorem shown above. All summands of the series are clearly positive, so it remains to show that there is an upper bound for the partial sums. We employ a similar trick as in showing the divergence of the harmonic series.^{[4]} We consider the first summands, which we split into suitable groups, such that each group can be assigned a bound. However, this time we choose upper bounds for the groups, as we eventually want to prove convergence. We use summands instead of because this eases finding a bound. At the end of a chain of inequalities we arrive at the geometric series, whose convergence criterion has been shown in this article.^{[5]}

Hence we have found an upper bound for the partial sum . Moreover, this upper bound is independent of , so it is not only an upper bound for a particular partial sum, but for all at the same time. We have thus shown that the subsequence is bounded. However, this is not sufficient to show boundedness of the *full* sequence of partial sums.

To address this problem, we need an auxiliary argument. Using the Archimedean property of and Bernoulli's inequality, we were able to show^{[6]} that for any positive integer there is another integer such that . In other words, any natural number is smaller than some power of two. Similarly, we can show that for any given we can also find some such that . As the sequence of partial sums is monotonically increasing, this implies . So any partial sum is bounded by one of the partial sums for which we have derived a *universal* bound above. We can conclude that this bound is a bound for *all* partial sums, such that the criterion is applicable to the series.

**Proof** (Convergence of the hyperharmonic series)

For any and , we have

This proves monotonicity of the sequence of partial sums.

Furthermore, for any there is some such that . This enables the chain of inequalities

Thus, the sequence of partial sums is also bounded. The monotonicity criterion then shows convergence of the hyperharmonic series for any .