Often, our goal is to show that a given sequence is convergent.
However, as we study convergent sequences, we would like
to establish various properties that they have in common.
The first theorem of this section is just such a result.
Suppose {an}{an} is a convergent sequence of real or complex numbers.
Then the sequence {an}{an} forms a bounded set.
Write L=liman.L=liman.
Let ϵϵ be the positive number 1.1.
Then, there exists a natural number NN such that
|an-L|<1|an-L|<1
for all n≥N.n≥N.
By the backward triangle inequality,
this implies that ||an|-|L||<1||an|-|L||<1 for all n≥N,n≥N, which implies that
|an|≤|L|+1|an|≤|L|+1 for all n≥N.n≥N.
This shows that at least the tail of the sequence is bounded by the constant |L|+1.|L|+1.
Next, let KK be a number larger than the finitely many numbers |a1|,...,|aN-1|.|a1|,...,|aN-1|.
Then, for any n,n,|an||an| is either less than KK or |L|+1.|L|+1.
Let MM be the larger of the two numbers KK and |L|+1.|L|+1.
Then |an|<M|an|<M for all n.n.
Hence, the sequence {an}{an} is bounded.
Note that the preceding theorem is a partial converse
to (Reference); i.e., a convergent sequence is necessarily bounded.
Of course, not every convergent sequence must be
either nondecreasing or nonincreasing, so that a full converse to
(Reference) is not true.
For instance, take z=-1/2z=-1/2 in part (1) of (Reference).
It converges to 0 all right, but it is neither nondecreasing nor nonincreasing.
- Suppose {an}{an} is a sequence of real numbers that converges to a number a,a,
and assume that an≥can≥c for all n.n.
Prove that a≥c.a≥c.
HINT: Suppose not, and let ϵϵ be the positive number c-a.c-a.
Let NN be a natural number corresponding to this choice of ϵ,ϵ,
and derive a contradiction.
- If {an}{an} is a sequence of real numbers for which liman=a,liman=a, and if a≠0,a≠0, then prove that an≠0an≠0 for all large enough
n.n.
Show in fact that there exists an NN such that
|an|>|a|/2|an|>|a|/2 for all n≥N.n≥N.
HINT:
Make use of the positive number ϵ=|a|/2.ϵ=|a|/2.
- If {an}{an} is a sequence of positive real numbers for which liman=a>0,liman=a>0,
prove that liman=a.liman=a.
HINT: Multiply the expression an-aan-a above and below by
an+a.an+a.
- If {an}{an} is a sequence of complex numbers, and
liman=a,liman=a, prove that
lim|an|=|a|.lim|an|=|a|.
HINT: Use the backward triangle inequality.
Suppose {an}{an} is a sequence of real numbers and that L=liman.L=liman.
Let M1M1 and M2M2 be real numbers such that
M1≤an≤M2M1≤an≤M2 for all n.n.
Prove that M1≤L≤M2.M1≤L≤M2.
HINT: Suppose, for instance, that L>M2.L>M2.
Make use of the positive number L-M2L-M2 to derive a contradiction.
We are often able to show that a sequence converges by comparing it
to another sequence that we already know converges.
The following exercise demonstrates some of these techniques.
Let {an}{an} be a sequence of complex numbers.
- Suppose that, for each n,n,|an|<1/n.|an|<1/n.
Prove that 0=liman.0=liman.
- Suppose {bn}{bn} is a sequence that converges to 0,0, and suppose that, for each n,n,|an|<|bn|.|an|<|bn|.
Prove that
0=liman.0=liman.
The next result is perhaps the most powerful technique
we have for showing that a given sequence converges to a given number.
Suppose that {an}{an} is a sequence of real numbers and that {bn}{bn} and {cn}{cn} are two sequences
of
real numbers for which
bn≤an≤cnbn≤an≤cn for all n.n.
Suppose further that limbn=limcn=L.limbn=limcn=L.
Then the sequence {an}{an} also converges to L.L.
We examine the quantity |an-L,||an-L,| employ some add and subtract tricks, and make the following computations:
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(1)So, we can make |an-L|<ϵ|an-L|<ϵ by making |cn-L|<ϵ/3|cn-L|<ϵ/3 and
|bn-L|<ϵ/3.|bn-L|<ϵ/3.
So, let N1N1 be a positive integer such that |cn-L|<ϵ/3|cn-L|<ϵ/3 if n≥N1,n≥N1,
and let N2N2 be a positive integer so that |bn-L|<ϵ/3|bn-L|<ϵ/3 if n≥N2.n≥N2.
Then set N=max(N1,N2).N=max(N1,N2).
Clearly, if n≥N,n≥N, then both inequalities |cn-L|<ϵ/3|cn-L|<ϵ/3 and |bn-L|<ϵ/3,|bn-L|<ϵ/3, and hence
|an-L|<ϵ.|an-L|<ϵ.
This finishes the proof.
The next result establishes what are frequently called the “limit theorems.”
Basically, these results show how convergence
interacts with algebraic operations.
Let {an}{an} and {bn}{bn} be two sequences of complex numbers with a=limana=liman and b=limbn.b=limbn.
Then
- The sequence {an+bn}{an+bn} converges, and
lim(an+bn)=liman+limbn=a+b.lim(an+bn)=liman+limbn=a+b.
(2) - The sequence {anbn}{anbn} is convergent, and
lim(anbn)=limanlimbn=ab.lim(anbn)=limanlimbn=ab.
(3) - If all the bnbn's as well as bb are nonzero, then the sequence
{an/bn}{an/bn} is convergent, and
lim(anbn=limanlimbn=ab.lim(anbn=limanlimbn=ab.
(4)
Part (1) is exactly the same as (Reference).
Let us prove part (2).
By Theorem 1, both sequences {an}{an} and {bn}{bn} are bounded.
Therefore, let MM be a number such that |an|≤M|an|≤M
and |bn|≤M|bn|≤M for all n.n.
Now, let ϵ>0ϵ>0 be given. There exists an N1N1 such that
|an-a|<ϵ/(2M)|an-a|<ϵ/(2M) whenever n≥N1,n≥N1,
and there exists an N2N2 such that |bn-b|<ϵ/(2M)|bn-b|<ϵ/(2M)
whenever n≥N2.n≥N2.
Let NN be the maximum of N1N1 and N2.N2.
Here comes the add and subtract trick again.
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(5)if n≥N,n≥N,
which shows that lim(anbn)=ab.lim(anbn)=ab.
To prove part (3), let MM be as in the previous paragraph, and let
ϵ>0ϵ>0 be given. There exists an N1N1
such that |an-a|<(ϵ|b|2)/(4M)|an-a|<(ϵ|b|2)/(4M) whenever n≥N1;n≥N1;
there also exists an N2N2 such that |bn-b|<(ϵ|b|2)/(4M)|bn-b|<(ϵ|b|2)/(4M) whenever
n≥N2;n≥N2; and there exists an
N3N3 such that |bn|>|b|/2|bn|>|b|/2 whenever n≥N3.n≥N3.
(See Exercise 1.)
Let NN be the maximum of the three numbers N1,N2N1,N2 and N3.N3. Then:
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(6)if n≥N.n≥N.
This completes the proof.
REMARK
The proof of part (3) of the preceding theorem may look mysterious.
Where, for instance, does this number ϵ|b|2/4Mϵ|b|2/4M come from?
The answer is that one begins such a proof by examining the quantity
|an/bn-a/b||an/bn-a/b| to see if by some algebraic manipulation one can discover
how to control its size by using the quantities |an-a||an-a| and |bn-b|.|bn-b|.
The assumption that a=limana=liman and b=limbnb=limbn
mean exactly that the quantities |an-a||an-a| and |bn-b||bn-b| can be controlled by requiring
nn to be large enough.
The algebraic computation in the proof above shows that
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(7)and one can then see exactly how small to make |an-a||an-a| and |bn-b||bn-b| so that |an/bn-a/b|<ϵ.|an/bn-a/b|<ϵ.
Indeed, this is the way most limit proofs work.
If possible, determine the limits of the following sequences
by using (Reference), (Reference), Theorem 3, and the squeeze theorem Theorem 2.
- {n1/n2}.{n1/n2}.
- {(n2)1/n}.{(n2)1/n}.
- {(1+n)1/n}.{(1+n)1/n}.
- {(1+n2)1/n3}.{(1+n2)1/n3}.
- {(1+1/n)2/n}.{(1+1/n)2/n}.
- {(1+1/n)2n}.{(1+1/n)2n}.
- {(1+1/n)n2}.{(1+1/n)n2}.
- {(1-1/n)n}.{(1-1/n)n}.
HINT: Note that
1-1/n=n-1n=1nn-1=1n-1+1n-1=11+1n-1.1-1/n=n-1n=1nn-1=1n-1+1n-1=11+1n-1.
(8) - {(1-1/(2n))3n}.{(1-1/(2n))3n}.
- {(n!)1/n}.{(n!)1/n}.
