Introductory Example[Bearbeiten]
Sometimes its necessary to speak about the "subsequence" of a sequence. Much like a subset is a "part" of a set, a subsequence is some part of a sequence -- all elements of a subsequence are also elements of the original sequence. A subsequence is in end effect constructed by removing some chosen terms of the original sequence. Regardless of how many terms are stripped from the original sequence, the resulting subsequence still has an infinite number of terms. For example, let's take the sequence
:
We are interested in a subsequence composed of every other term of the original sequence
. This subsequence arises from either removing all terms with an even index or removing all terms with an odd index. If, for example, we remove all terms with odd index, we get the following schematic:
This gives rise to a subsequence that is constant
.
Mathematical Description[Bearbeiten]
How can subsequences be denoted? First let's look at the indices of the sequence terms that we want to keep in the subsequence:
Now we want to find a sequence
that describes these indices. In the above example we consider even indices. So the sequence can be written as
:
We substitute this sequence into
. From there we get the subsequence
:
First we will build the sequence
of relevant indices of the subsequence. We will set this subsequence into the original sequence
for
so that we get the sequence
.
In our example we have
. So we substitute
for
in
. Then we obtain the subsequence
.
This concept is important for analysis since it is used to characterize so-called "limit points." However, these will not be properly defined and discussed until the next chapter.
Exercise (Subsequences)
Find five different subsequences of the sequence
.
Solution (Subsequences)
The sequence
has infinitely many subsequences. Five examples would be
Convergence of Subsequences[Bearbeiten]
For subsequences we have the following important theorem:
Theorem (Convergence of Subsequences)
Let
be a sequence. It converges if and only if every subsequence converges. The limit of the sequence coincides with the limits of the subsequences.
Proof (Convergence of Subsequences)
To prove the equivalence
we have to prove the following two implications:
- If
converges to
, then every subsequence of
converges to
, as well.
- If every subsequence of
converges to
, then
converges to
, as well.
We can split the proof into these two parts, since the statement
is equivalent to the two implications
.
Proof step: Convergence of the sequence implicates the convergence of all subsequences - to the same limit.
Let
be a sequence converging to
.Now, we need to prove that all subsequences of
converge to
, as well.
So, let
be a subseqeunce of
. We would like to show that
converges to
. Let
be given arbitrarily. Since
is the limit of
, there must be an index
, such that for all
the inequality
holds.
Since the sequence
is by definition strictly decreasing , there will be
for all
. Therefore,
for all
, since from
and
we conclude
. Hence, there is also
for all
.
We have just proven that for any given
there is an
with
for all
. But this is just the definition for "
converges to
" . Since any subsequence
can be chosen in this place, the proof holds for all subsequences of
.
Proof step: Convergence of all subsequences to the same limit implies convergence of the sequence.
Example (Convergence of subsequences)
Since
is a null sequence, the same will hold for the two limits
Hint
The above theorem directly implies that a convergent sequence
does not change its limit, if a finite number of elements are canceled. Removing a finite amount of elements will render a subsequence
of
. And we just showed that the limits of those two sequences coincide.
The above theorem directly implies:
Theorem (Divergence in case of a diverging subsequence)
If a subsequence diverges, the original sequence must also diverge.
Check your understanding: Why does the above theorem imply that the original sequence has to diverge, if any subsequence does so?
We know: If a sequence converges, then all of its subsequences also have to diverge. If there was a convergent sequence for which we could find a divergent subsequence, we would have a contradiction to the theorem. So if a subsequence diverges, we know that the original sequence must have been convergent.
Application: convergence of mixed sequences[Bearbeiten]
In the chapter „Beispiele und Eigenschaften von Folgen“ we have seen how to compose to sequences
and
to a "mixed sequence"
. This mixture is defined as
That means, the sequence
is composed out of the two subsequences
and
.
We may now ask how the convergence of the mixture
relates to the convergences of its two constituents
and
.
In order for
to converge, two conditions must be satisfied:
- First, both subsequences
and
have to converge, as we know that for convergent sequences, all subsequences converge.
- Second, the limits of
and
must be identical. This is because if
converges, then all of its subsequences must tend to the same limit.
If one of these two conditions is not satisfied, the mixed sequence
must diverge. But are the two conditions also sufficient for convergence of the mixed sequence? Indeed, they are!
We will now proof this. The limit of the mixed sequence must the coincide with the two limits of the subsequences.
Theorem (Convergence of mixed sequences)
Let

and

be two sequences and

a limit candidate. Let the mixed sequence

be defined by
. It converges to

, if and only if the sequences

and

both converge to

.
Proof (Convergence of mixed sequences)
Proof step: If both
and
converge to
, then also
converges to
.
As both
and
converge to
, the following two statements hold:


Since
and
there is:


Let now
be given arbitrarily. The above statements imply that there is a threshold number
(for b) with
for all
. In addition, there is a threshold
(for c) with
for all
. We choose the maximum of both thresholds
. Let
be a number above this threshold. If
is odd, then
for some
. Since
, there is especially
and hence
In case
is even, we know that
for some
. As
, there is
and therefore
In any case,
for all
. This establishes convergence
.
Check your understanding: In the above proof, we chose
. Why couldn't we have chosen
?