Real Analysis
Supplemental: How to write an – N proof.
The purpose of this document is to talk you through some of the types of proofs you will
need.
How to prove that the limit of a particular sequence is a particular value.
1
You know already that lim 3 −
= 3. Let’s talk about the proof. The lines in red are
n→∞
n+1
comments about the proof rather than actual parts of the proof. They will help you adapt
this proof to a new setting.
4
= 3.
Claim: lim 3 −
n→∞
n+1
Proof. Consider any > 0. In general, proofs of claims involving limits start with this line
Now we do some scratch work. We need to decide what choice of N will allow us to deduce
4
that |3 − n+1
− 3| < from the assumption that n > N .
4
− 3| < .
We want: |3 − n+1
1
Doing algebra: 4| n+1 | < .
1
n+1 < 4
n > 4 − 1
We should try N = 4 − 1
Let N = 4 − 1. Consider any n ∈ N such that n > N .
Now see if you can invert the analysis in red so it goes in the right order.
By assumption, n > N = 4 − 1.
Adding 1 to both sides: n + 1 > 4 .
1
Inverting n+1
< 4 .
1
Multiply by 4 n+1 < .
1
1
− n+1
| <.
n+1 is positive | 1
− 3| < .
Adding and subtracting 3 | 3 − n+1
4
Thus, we see that for any > 0 there is a N > 0 such that if n > N then |3 − n+1
− 3| < .
4
Thus, lim 3 −
= 3.
n→∞
n+1
As a reminder, the actual proof does not include any of the red text.
Using a claim about the limit of one sequence to make a conclusion about
another limit.
These proofs will usually proceed by using the assumed limits to get a choice N .
Claim: Let (an ) be a convergent sequence. Let L = lim an . For any c ∈ R c·L = lim c·an .
n→∞
n→∞
Proof. Consider any > 0. almost every single one of your proofs will start this way. We
need to get an N . Use the assumption that L = lim an . By taking N big enough and n > N
n→∞
we can guarantee that |an − L| < ? for any positive ?. Try to transform |c · an − c · L| < to
|an − L| < something:
1
2
We want: |c · an − c · L| < factor out the |c| and divide |an − L| < |c|
we should use |c| = ?
Does anything go wrong when c = 0? We need a special case for c = 0. We will do that at
the end of the proof.
Case 1: Assume that c 6= 0. Then since |c|
> 0 there is an N such that for all n > N ,
|an − L| <
.
|c|
Multiply both sides by |c|,
|c||an − L| < .
|c · an − c · L| < .
Thus, for all > 0 there is an N such that if n > N then |c · an − c · L| < . This is precisely
the definition of the desired claim that c · L = lim c · an .
n→∞
Claim: Let (an ) and (bn ) be convergent sequences. Let L = lim an and M = lim bn .
n→∞
n→∞
Then L + M = lim an + bn .
n→∞
Proof. Consider any > 0.
The proof is similar, except that we now have two assumptions, to be precise, by taking n
greater than some N1 we can get |an − L| < ? (we are free to chose ?), by taking n greater
than some other possible different N2 we get |bn − M | < 4 (again 4 can be chosen.)
How can we get |an + bn − L − M | < out of these inequalities? Rearrange and use the
triangle inequality.
|an + bn − L − M | = |an − L + bn − M | ≤ |an − L| + |bn − M | < ? + 4
We get to choose ? and 4. How about if they are both 2 ? Now we have a strategy!
Since L = lim an , there exists some N1 such that for all n > N1
n→∞
|an − L| < .
2
Now we can use(1) to refer to this inequality later. Similarly, since M = lim bn , there exists
n→∞
some N2 such that for all n > N2
(2)
|bn − M | < .
2
We need an N that is at least as big as N1 and as big as N2 . Let N = max(N1 , N2 ). Then
N ≥ N1 and N ≥ N2 . Consider any n > N . Then inequalities (1) and (2) follow.
(1)
By the commutativity of addition, |an + bn − L − M | = |an − L + bn − M |.
By the triangle inequality,
≤ |an − L| + |bn − M |.
By inequalities (1) and (2),
≤ 2 + 2 = .
Thus, for all > 0 we have found a choice of N such that if n > N then |an +bn −L−M | < .
Thus we can conclude that L = lim an .
n→∞
Sometimes, the parameter n changes.
Claim: Suppose that (an ) is a convergent sequence and L = lim an . Then L = lim a2n .
3
Proof. Consider any > 0. Use the assumption. Since L = lim an ,there is an N0 such that if
m > N0 then
|am − L| < .
(3)
We use N0 since this is not quite the N we need. Notice that if 2n > N0 then we will be able
to conclude that |a2n − L| < . Perhaps we should try N = N20 ?
Let N = N20 and n > N = N20 . Then 2n > N0 , so that taking m = 2n in (3) produces
the result that |a2n − L| < . Thus, for any > 0 there is an N such that if n > N then
|a2n − L| < and we conclude L = lim a2n .
Here is a harder one. Try to use the ideas in this handout to recover the proof which
appears in the book.
Claim: Let (an ) and (bn ) be convergent sequences. Let L = lim an and M = lim bn .
n→∞
n→∞
Then L · M = lim an · bn .
n→∞