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]
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.
Exercise (Monotony criterion for sequences)
Show, using the monotony criterion, that the sequence
with
converges
.
Solution (Monotony criterion for sequences)
Step 1: Monotony of
.
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
.
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:
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.
.
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.