This page is about subsequences of a sequence. A subsequence of a sequence is an infinite collection of numbers from in the same order that they appear in that sequence.
The main theorem on subsequences is that every subsequence of a convergent sequence converges to the same limit as . This (together with the Theorem on Uniqueness of Limits) is the main tool in showing a sequence does not converge.
It is harder to use subsequences to prove a sequence does converge and much easier to make mistakes in this direction. Just knowing a sequence has a convergent subsequence says nothing about the convergence of . We discuss these topics with an example in another web page.
If is a sequence, such as we may obtain a subsequence by selecting some of the
terms in the sequence. The only rules are that the order
of the terms in the original sequence must be preserved
and we must select an infinite number of terms. For example,
from the above sequence we may select
and so on. This idea of
subsequence is rather simple
and very natural. Unfortunately there are some difficulties
with notation that tends to obscure the main idea and can make
the manipulation of subsequences a little more tricky.
Definition of subsequence.
Suppose is a sequence and is an increasing function (i.e., ) then and define a new sequence
The sequence is called a subsequence of .
For the sequence given by we obtain the subsequence by taking as our increasing function , since .
The ideas of
selecting infinitely many terms from a sequence
and using an increasing function to enumerate a selection of terms
are equivalent. If we have an infinite set of terms we can enumerate
them with a function . Conversely, an increasing function
defines a selection of terms with indices in the image of .
The following lemma is useful in switching between these two views.
Let be increasing. Then for all .
By induction on . Suppose for the sake of this proof that (a similar argument starting with one works if you prefer not to be a natural number) and let the induction hypothesis be . Then and so , so the base case is established. Now inductively assume and observe that as is increasing. So by our hypothesis, and the induction step is proved.
Theorem on Subsequences.
Suppose as and is a subsequence of defined by . Then as .
Let be arbitrary.
Let satisfy .
Let be arbitrary.
Then by the lemma.
Part of the power of this theorem is that the subsequence converges to the same limit. This will be combined with the Theorem on Uniqueness of Limits to allow us to prove non-convergence results.
We use these ideas to re-prove the assertion that the sequence doesn't converge. Define subsequences and , and observe that for all and for all , so and as , as these are constant sequences. Now suppose (to get a contradiction) that converges to some limit . Then by the Theorem on Subsequences both and as these are subsequences of . By the Theorem on Uniqueness of Limits, and . But this implies which is false.
Consider the sequence defined by . This is proved to be non-convergent as follows.
First a subsequence is selected consisting of all for which . It may not be immediately obvious that there are infinitely many such but this is easily proved as for each the integer part of is such an . Note for such that .
Next a subsequence is selected consisting of all for which . Again, note that for each the integer part of is such an . Note also that for such that .
Finally, suppose . Then by the Theorem on Subsequences both and . But by the Theorem Giving Bounds on Limits we have as is the limit of . Also as is the limit of . But this is impossible as .
Be careful to note the and here! In particular this argument wouldn't have worked if we have simply shown that for all for then it might have been possible that .