Monotony criterion – Serlo

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In this chapter, we will prove that monotonous and bounded functions converge. So if you have a bounded sequence, which in addition is monotonous, you instantly know that it converges. There is no need ot make complicated bounds within the -definition. You do not even have to know what the limit is!

This especially turns out to be useful for recursive sequences. For those sequences, there is often no explicit form available which makes it hard to guess, what a limit may be or what the difference of a sequence element to a potential limit is.

Convergence of monotonous and bounded sequences[Bearbeiten]

Convergence of monotonous and bounded functions. (YouTube-Video in German, by the channel Quatematik)

Theorem (Monotony criterion for sequences)

Every monotone and bounded sequence in the real numbers converges.

It is useful to get a visual representation of this theorem: Take a piece of paper and try draw a sequence of points, which is monotonously increasing and bounded, but still diverges. How could it possibly diverge. Maybe to ? This is certainly not possible because the sequence is bounded. The only option left for a divergence is that the function "jumps" between different points. But now, monotony tells us that , so the sequence goes always up and never down. Therefore, "jumping" is also not allowed. The sequence either goes up until it reaches the upper bound (and hence converges to it), or it gets "stuck" before reaching the bound and converges to some smaller number (below the bound). So intuitively, there should be no way for the sequence to converge. Let us make this statement watertight by formulating it as a mathematical theorem:

Proof (Monotony criterion for sequences)

The sequence

We consider monotonously increasing sequences. The proof for monotonously decreasing sequences works analogously. Let be a monotonously increasing and bounded function. We need to find a potential limit and then prove that the sequence actually converges to it (using the Epsilon-definition). To see, how this works, we first consider the sequence as an example:

The limit is 1. And obviously, the sequence elements approach 1, but never reach it. So 1 is a supremum to the sequence elements . For a general monotonous and bounded sequence , we may also assume that it converges towards its supremum, since for convergence to a smaller number, the sequence would have to "go down again". So let us prove that converges to

This supremum must exist, since is bounded from above.

Now, let us turn to the convergence proof: Let be given. Depending on this we need to find an such that for all .

As is a supremum, we know that there is an , which is greater than . If this was not the case, would be an upper bound to the sequence (which can not be, since as a supremum is the smallest upper bound). Hence, there is

Which implies

So is close enough to . But are all with also closer that to ? If the sequence was not monotonous, there would be no warranty for that and it might be "divergent by jumping around"! But our sequence is monotonously increasing, so all with must be bigger or equal to . In addition, , since is an upper bound to the sequence. For we hence have

so

This finishes the proof of being the limit of and the sequence converges. For monotonously decreasing sequences, the proof works analogously, choosing the infimum instead of the supremum.

An example[Bearbeiten]

Exercise (Monotony criterion for sequences)

Show, using the monotony criterion, that the sequence with

converges .

Solution (Monotony criterion for sequences)

Step 1: Monotony of .

In order to apply the criterion, we need to make sure that is monotonous. We have a sequence of sums and the difference between subsequent elements is

I.e. for all and is monotonously decreasing.

Step 2: Boundedness of

Since decreases monotonously, we need to show that it is bounded from below This is very easy to see: all summands are , so the sequence elements are also . In order to find out what the potential limit is, we can try to even find a better (= higher) lower bound:

So (and also its limit) is bounded from below by .

Hence, the monotony criterion can be applied and converges.

Hint

Later, we will actually be able to show that the limit is .

Nested intervals[Bearbeiten]

There is a useful implication for nested intervals.

Recap: A sequence of nested intervals is a sequence with the following properties:

1. All intervals are subsets of their precursors:

2. For all there is an interval smaller than :

In that case, there is always exactly one real number being included in all intervals. This statement can be shown using the monotony criterion:

We take a look at the two sequences and , i.e. upper and lower bounds of the intervals.

  • Since there is
, i.e. , and , i.e.

So is monotonously increasing and monotonously decreasing.

  • Since and , the sequence is bounded from above by and conversely, is bounded from below by .

The monotony criterion hence implies that and converge. The limit of both is even equal and exactly this one number being in all intervals:

Since is a sequence of nested intervals, there is

As we also have for all .

So the difference of and converges to0:

Using the limit theorems, we hence get that and have the same limit. This limit is

So , i.e. and is included in all intervals. All other real numbers or with can not be inside all intervals, since they either exceed as an upper bound or as a lower bound. More precisely, for , there is a and is not inside the interval . An analogous problem appears for . So there is exactly one real number included in all intervals.

We recap those considerations in a theorem:

Theorem

Let be a sequence of nested intervals. Then, and both converge to one limit

and is the only real number with

Application: Euler's number[Bearbeiten]

Computing Euler's number e by nested intervals

We consider the sequence of intervals with and .

These can be shown to be nested intervals, which we will do in the following. In that case, both and converge. At first,

So is well-defined.

For a sequence of nested intervals, we need to establish two properties. At first, for all : (intervals are included in each other). This is done in two steps:

  • The lower bounds are monotonously increasing, i.e. . This is done by showing , using Bernoulli's inequality:
  • The upper bounds are monotonously increasing, i.e. .

Exercise

Prove this.

How to get to the proof?

The proof works analogously to the case above with :

Proof

Since is monotonously increasing and decreasing, the intervals are included in each other, i.e. . So we have established the first property for nested intervals.

The second property is that interval sizes go to 0:

This works by bounding from above.

But now, , so

There is . If we are given any and choose a corresponding with , then

So the second property is also established and is indeed a sequence of nested intervals. The number included in all intervals is called Euler's number . The sequence of nested intervals can be used for making estimates, what is. For instance, . For greater , one would get even more digits, e.g. .

With the theorem above, there is

In the series chapter, we will show that with denoting the exponential series.