Download Sequences and limits

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4.1
Sequences
Definition and basic properties
By a real valued sequence we will mean
(an )n∈N = (a1 , a2 , a3 , . . .)
where for all i ∈ N, ai ∈ R. The formal way of defining this would be
as a function from the natural numbers to the real numbers. Note that ai
represents the ith element of the sequence and (an )n∈N often shortened to
(an ) represents the whole sequence. The usual way to define a sequence will
be by specifying a formula which an satisfies for all n ∈ N.
Example. Let (an )n∈N be the sequence where an = (−1)n for all n ∈ N. In
this case
(an )n∈N = (−1, 1, −1, 1, −1, . . .).
We would have that a7 = (−1)7 = −1.
Definition 4.1. A sequence of real numbers (an )n∈N is bounded if there
exists K > 0 such that for all n ∈ N, |an | ≤ K. We say that (an )n∈N is
bounded below if there exists K ∈ R such that an ≥ K for all n ∈ N and
bounded above if there exists K ∈ R such that an ≤ K for all n ∈ N.
Example. The sequence (an )n∈N given by an = (−1)n is bounded since for
all n ∈ N we have that |an | = 1. On the other hand the sequence (bn )n∈N
where bn = n is not bounded. To see this we need to show that for any
K > 0 we can find n ∈ N such that |bn | > K. So we let K > 0 and note
that by the Archimedean principle we can find n ∈ N where n > K and so
|bn | = n > K. Note that the sequence (bn )n∈N is bounded below by 1.
4.2
Convergent sequences
We now give the most important definition in this unit.
Definition 4.2. A sequence (an )n∈N of real numbers is convergent to α ∈ R
if for all ε > 0 there exists N ∈ N such that for all n ∈ N, if n ≥ N then we
have that
|an − α| ≤ ε.
When a sequence of real numbers (an )n∈N is convergent to α ∈ R we say
that the sequence is convergent and write limn→∞ an = α.
This means that in order to show that a sequence (an )n∈N converges to
a real number α you need to show that whichever ε > 0 you are given you
can find N ∈ N such that whenever you have a natural number n ≥ N you
have that |an − α| ≤ ε.
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Example. We start with a trivial example. If we let c ∈ R and an = c for
all n ∈ N then (an )n∈N is convergent with limn→∞ an = c. To see this let
ε > 0 and choose N = 1. We then have that if n ∈ N and n ≥ 1,
|an − c| = 0 ≤ ε.
Example. Let an = n1 for all n ∈ N. We will show that limn→∞ an = 0.
Let ε > 0 and by the Archimedean principle choose N ∈ N with N > 1ε . We
have that for any n ∈ N with n ≥ N ,
|an − 0| =
1
1
≤
≤ ε.
n
N
Thus we have shown that for any ε > 0 there exists N ∈ N such that for
all natural numbers n ≥ N we have |an − 0| ≤ ε. Therefore (an )n∈N is
convergent with limn→∞ an = 0.
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Example. Let an = nn2 −1
for all n ∈ N. We will show that limn→∞ an = 1.
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Let ε > 0 and by the Archimedean principle choose N ∈ N such that N > √2ε .
We have that for any n ≥ N ,
2
2
n − 1
n − 1 − (n2 + 1) 2
2
2
=
|an −1| = 2
− 1 = ≤ 2
≤ 2 ≤ ε.
2
2
n +1
n +1
n +1
n +1
N
Therefore (an )n∈N is convergent with limn→∞ an = 1.
Remark. In the second example before writing it down as written above it is
necessary to do some work to figure out how to choose N . This is standard
with these arguments while they should be written in the form ‘Let ε > 0 and
choose N = · · · ’ it is usually necessary to do some rough work to figure out
how to choose N before writing this down.
We can use the definition of a limit to say exactly what we mean by a
sequence being convergent or divergent.
Definition 4.3. Let (an )n∈N be a real valued sequence we say that (an )n∈N
is convergent if and only if there exists α ∈ R where limn→∞ an = α. We
say that a sequence (an )n∈N is divergent if it is not convergent.
To show that a sequence is divergent means that we have to show it is
not convergent to any α ∈ R. So by negating the definition of a limit we have
that a sequence (an )n∈N is divergent if for any α ∈ R there exists ε > 0 such
that for all N ∈ N there exists n ∈ N with n ≥ N where |an − α| > ε. This
is quite a clunky process and in practice we won’t often use this definition
directly (instead we show that the sequence does not satisfy some property
that all convergent sequences satisfy). However we will now give a couple of
examples working directly from the definition.
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Examples. Let (an )n∈N be the sequence given by an = (−1)n for all n ∈ N.
We will show that (an )n∈N is divergent. Let α ∈ R and choose ε = 1/2.
We have that for any N ∈ N that |aN +1 − aN | = 2. So 2 = |aN +1 − aN | ≤
|aN +1 − α| + |aN − α|. Thus we must have that either |aN +1 − α| > 12 = ε
or |aN − α| > 21 = ε. Either way we have shown that for all α ∈ R for ε = 21
and for any N ∈ N that we can find n ∈ N with n ≥ N and |an − α| > ε
(either take n = N or n = N + 1.)
√
Let (bn )n∈N be the sequence given by bn = n for all n ∈ N. We will show
that (bn )n∈N is divergent. Let α ∈ R and choose ε = 1. By the Archimedean
property we can find k ∈ N with k > (α + 1)2 . Thus if we let√N ∈ N and
√
take n = max{N, k} then we have that n ≥ N and bn = n ≥ k > α + 1.
Thus |bn − α| > 1 and so (bn )n∈N is divergent.
4.3
Basic properties of convergent sequences
In this section we look at some simple consequences of a sequence being
convergent. First of all a convergent sequence is bounded.
Theorem 4.4. If (an )n∈N is a convergent sequence of real numbers then
(an ) is bounded
Proof. Since (an ) is convergent we know there exists α ∈ R with limn→∞ an =
α. Taking ε = 1 in the definition of convergence this means there exists N ∈
N such that for all n ≥ N , |an −α| ≤ 1. Now let K = max{|ai | : 1 ≤ i ≤ N }.
For n ≤ N we have that by definition |an | ≤ K. If n ≥ N we have by the
triangle inequality
|an | = |an − α + α − aN + aN | ≤ |an − α| + |α − aN | + |aN | ≤ K + 2.
Therefore for all n ∈ N we have that |an | ≤ K +2. Thus (an )n∈N is bounded.
This means that any unbounded sequence must be divergent. However it
is possible to have a bounded sequence which is not convergent. For example
we have already seen that an = (−1)n for all n ∈ N is divergent but |an | = 1
for all n ∈ N so we can see that the sequence is bounded.
Secondly the limit of a sequence may only take one value
Proposition 4.5. If (an )n∈N is a convergent sequence and a, b ∈ R with
limn→∞ an = a and limn→∞ an = b then a = b.
Proof. Suppose that limn→∞ an = a and limn→∞ an = b. We will show that
for all ε > 0, |a − b| ≤ ε and thus a = b. Let ε > 0. We can choose N1 ∈ N
such that for all n ∈ N with n ≥ N1 we have that |an − a| ≤ ε/2. We
can also choose N2 ∈ N such that for all n ∈ N with n ≥ N2 we have that
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|an − b| ≤ ε/2. Now take N = max{N1 , N2 } and note that by the triangle
inequality,
|a − b| = |a − aN + aN − b| ≤ |a − aN | + |aN − b| ≤ ε/2 + ε/2 = ε.
Thus for all ε > 0 we know that |a − b| ≤ ε and so a = b.
We now give a couple of equivalent definitions of convergence which can
be useful.
Proposition 4.6. Let (an )n∈N be a sequence of real numbers and α ∈ R.
The following statements are equivalent.
1. limn→∞ an = α.
2. There exists K > 0 such that for all ε > 0 there exists N ∈ N such
that for all n ∈ N with n ≥ N ,
|an − α| ≤ Kε.
3. For all ε > 0 there exists N ∈ N such that for all n ∈ N with n ≥ N ,
|an − α| < ε.
Proof. To show all the statements are equivalent we will show that 1 implies
2 that 2 implies 3 and then 3 implies 1.
To see that 1 implies 2 we simply need to take K = 1. Then since (an )
is convergent there will exist N ∈ N such that for all n ≥ N , |an − α| ≤ Kε.
To see that 2. implies 3. Since 2. is satisfied we know there exists K > 0
such that for all ε > 0 there will exist N ∈ N such that for all n ≥ N ,
|an − α| ≤ Kε. So we will fix such a K > 0 and let ε > 0. We then
choose N ∈ N such that for all n ≥ N , |an − α| ≤ K(εK −1 )/2 = ε/2 (since
K −1 ε > 0). Thus for all n ≥ N we have that |an − α| < ε and so 3. is
satisfied.
To see that 3. implies 1. is immediate.
We now state a theorem which is extremely useful for finding the limits
of certain sequences without working directly from the definition.
Theorem 4.7 (Sandwich rule). Let (an )n∈N , (bn )n∈N , (cn )n∈N be real valued
sequences and α ∈ R. If limn→∞ an = limn→∞ cn = α and for all n ∈ N
we have that an ≤ bn ≤ cn then we have that (bn )n∈N is convergent and
limn→∞ bn = α.
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Proof. Let ε > 0 and choose N ∈ N such that for all n ≥ N , |an − α| ≤ ε/3
and |cn −α| ≤ ε/3. We then have that for all n ≥ N by the triangle inequality
and the fact that an ≤ bn ≤ cn (which means that |bn − an | ≤ |cn − an |),
|bn − α| = |bn − an + an − α| ≤ |bn − an | + |an − α|
≤ |cn − an | + |an − α| ≤ |cn − α| + |α − an | + |an − α|
= |cn − α| + 2|an − α| ≤ ε.
Remark. Similarly we also have that for two sequences (an )n∈N and (bn )n∈N
that if an ≤ bn for all n ∈ N, limn→∞ an = a and limn→∞ bn = b then a ≤ b.
See the 4th exercise sheet.
We complete the section with a couple of examples using the sandwich
rule.
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Example. If we take (bn )n∈N to be the sequence with bn = 5n+6
for all
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1
n ∈ N we have that for all n ∈ N 0 ≤ 5n+6 ≤ n . We take (an ) to be
the sequence with an = 0 for all n ∈ N and (cn ) to be the sequence with
cn = n1 for all n ∈ N. We then have that an ≤ bn ≤ cn for all n ∈ N and
limn→∞ an = limn→∞ cn = 0. Thus by the sandwich rule limn→∞ bn = 0.
Corollary 4.8. For any k ∈ N, the sequence (an )n∈N defined by an = n−k
for all n ∈ N is convergent with limn→∞ an = 0.
Proof. Let k ∈ N we then have that 0 ≤ n−k ≤ n−1 and since limn→∞ 0 = 0
and limn→∞ n−1 = 0 the result follows by the sandwich rule.
4.4
Arithmetic properties of limits
Always working directly from the definition of a limit can be a long process.
However fortunately there are several results which we can use to make
things a much quicker process. We have already seen one example with the
sandwich rule. Here is another result which will help us do this.
Theorem 4.9. Let (an )n∈N , (bn )n∈N be real valued sequences and let a, b, c ∈
R. Suppose that limn→∞ an = a and limn→∞ bn = b. We have:
1. Sum rule: limn→∞ an + bn = a + b,
2. Product rule: limn→∞ an bn = ab,
3. Scalar product rule: limn→∞ can = ca,
4. Quotient rule: If b 6= 0 and bn 6= 0 for all n ∈ N then limn→∞
a
b.
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an
bn
=
Proof. Sum rule: Let ε > 0. Since we have that limn→∞ an = a and
limn→∞ bn = b we can choose N ∈ N such that if n ≥ N then |an − a| ≤ ε/2
and |bn − b| ≤ ε/2. Now if n ≥ n by the triangle inequality we have
|an + bn − (a + b)| ≤ |an − a| + |bn − b| ≤ ε/2 + ε/2 = ε.
Thus limn→∞ an + bn = a + b.
Product rule: Since (an ) and (bn ) are both convergent we know that they
are both bounded. So there exists K > 0 such that for all n ∈ N, |bn | ≤ K
and |a| ≤ K. Now let ε > 0 and choose N ∈ N such that if n ≥ N
ε
ε
|an − a| ≤ 2K
and |bn − b| ≤ 2K
. We then have that if n ≥ N by the triangle
inequality
|an bn − ab| = |an bn − abn + abn − ab| ≤ |an bn − abn | + |abn − ab|
= |bn ||an − a| + |a||bn − b| ≤ K(|an − a| + |bn − b|) ≤ ε.
Scalar product rule If we take (cn )n∈N to be the sequence with cn = c
for all n ∈ N then limn → ∞cn = c and so by the product rule
lim can = lim cn an = ca.
n→∞
n→∞
Quotient rule We will let cn be the sequence with cn = b1n for all n ∈ N.
We will first show that limn→∞ cn = 1b . Since b 6= 0 and limn→∞ bn = b
we can find N ∈ N such that for all n ∈ N with n ≥ N we have that
|bn − b| ≤ |b|/2 and so |bn | ≥ |b|/2. So we will let ε > 0 and choose N such
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that for all n ≥ N |bn | ≥ |b|/2 and |bn − b| ≤ εb2 . Thus by the triangle
inequality we have for n ≥ N ,
1
− 1 = bn − b = |bn − b| ≤ |bn − b| ≤ ε.
bn
b
bn b |bn b|
b2 /2
Thus limn→∞ cn = 1b . Since we can write
rule that limn→∞ abnn = ab .
an
bn
Example. Let (an )n∈N satisfy that an =
show that limn→∞ an = 25 . We have that
an =
= an cn it follows by the product
5n3 +6n+4
2n3 +n2 +9
for all n ∈ N. We will
5n3 + 6n + 4
5 + 6n−2 + 4n−3
=
.
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2
2n + n + 9
2 + n−1 + 9n−3
We have that by Corollary 4.8 limn→∞ n−k = 0 for all k ∈ N and we know
that limn→∞ c = c for all c ∈ R. So by the scalar product rule and the sum
rule limn→∞ 5 + 6n−2 + 4n−3 = 5 and limn→∞ 2 + n−1 + 9n−3 = 2. Moreover
2 + n−1 + 9n−3 > 0 for all n ∈ N so by the quotient rule limn→∞ an = 5/2.
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Remark. The example above illustrates a standard method when working
with this type of sequence. That is when dealing with a sequence defined
as a quotient, look for a dominant term (by this we mean the term in the
numerator or dominator which we expect to grow at the fastest rate). You
should then divide both numerator and denominator by this dominant term.
4.5
Divergent sequences
We will now look at some special cases of divergent sequences. These are
sequences which are divergent to ±∞.
Definition 4.10. Let (an )n∈N be a real valued sequence. We say that (an )
is divergent to ∞ and write limn→∞ an = ∞ if and only if for all K ∈ R
there exists N ∈ N such that for all n ≥ N , an > K. We say that (an ) is
divergent to −∞ and write limn→∞ an = −∞ if and only if for all K ∈ R
there exists N ∈ N such that for all n ≥ N , an < K.
Examples. Consider the sequence (an )n∈N where an =
We will show that limn→∞ an = ∞. First of all
n2 +1
n+4
for all n ∈ N.
n2 + 1
1 + n−2
1
n
= −1
≥ −1 = .
−2
n+4
n + 4n
5n
5
We now fix K ∈ R and note that by the Archimedean principle we can find
N ∈ N with N > 5K. Then for all n ≥ N ,
n
N
n2 + 1
≥ ≥
> K.
n+4
5
5
2
+1
Thus limn→∞ nn+4
= ∞.
Now consider the sequence (bn )n∈N with bn = (−1)n n2 for all N ∈ N we
will have that bn is nether divergent to ∞ or −∞, however |bn | is divergent
to infinity (limn→∞ |bn | = ∞). To justify this note that if n is odd then
bn ≤ 0 but if n is even then bn ≥ 0. Thus to see that (bn ) is not divergent
to infinity take K = 0 then for any N ∈ N either bN ≤ 0 or bN +1 ≤ 0 so
it is not possible to find N such that for all n ≥ N , bn > 0 so (bn ) is not
divergent to infinity. Similarly we can show (bn ) is not divergent to −∞.
However |bn | = n2 and we can easily show this is divergent to infinity.
Proposition 4.11. Let (an )n∈N be a real valued sequence with an 6= 0 for
any n ∈ N but limn→∞ an = 0. Let bn = a−1
n for all n ∈ N. We have that:
1. limn→∞ |bn | = ∞.
2. If an > 0 for all n ∈ N then limn→∞ bn = ∞.
3. If an < 0 for all n ∈ N then limn→∞ bn = −∞.
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Proof.
1. Let x ∈ R and choose ε = |x|+1
(this is so that ε−1 > |x|).
Since limn→∞ an = 0 we can choose N ∈ N such that for all n ≥ N ,
|an | ≤ ε. Thus for n ≥ N ,
|bn | =
1
1
≥ = |x| + 1 > |x| ≥ x.
|an |
ε
Thus limn→∞ |bn | = ∞.
2. We simply need to note that in this case bn = |bn | and so the result
follows by the previous part.
3. We leave this case as an exercise (you simply need to show that if
limn→∞ xn = ∞ then limn→∞ −xn = −∞.)
Remark. Several other properties of such sequences can be proved in similar
ways. There are quite a few examples of this on the 5th exercise sheet.
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