﻿ Example Proof Limit Using Explicit N

Projects

Back to Real Analysis

# Example Proof: Using the Definition of a Limit With Unknown Sequences

Using the Definition of Convergence

an converges to limit, L ⇔ ∀ 𝜀 > 0, ∃ N ∈ ℕ s.t. n ≥ N ⇒ |an - L| < 𝜀

show that if an converges to limit, A and bn converges to limit, B,
then an + bn converges to A + B.

Note:
1. Unlike the previous example, we do not know the form of the sequences or the limits.
So we will not be able to find a functional form of the "N" in the definition of convergence.

2. This is somewhat parallel to algebra, in which we did arithmetic without numbers (we used variables).
Here, we prove things about limits, without using actual functions.

## Thought Process:

In order to show the desired result, we need to show

∀ 𝜀 > 0, ∃ N ∈ ℕ s.t. n ≥ N ⇒ |(an + bn) - (A + B)| < 𝜀

The definition has in if and only if.
In the previous example, we found N with the required properties and concluded that the sequence converged.
We used: "∃ N s.t. ..." ⇒ the sequence converged to a limit.

We cannot use the same approach since the sequences are unknown, but we can try using the definition in the other direction.
"a sequence converged to limit" ⇒ ∃ N s.t. ..."

 an converges to A ⇒ ∀ 𝜀a > 0, ∃ Na(𝜀a) ∈ ℕ s.t. n ≥ Na(𝜀a) ⇒ |an - A| < 𝜀a bn converges to B ⇒ ∀ 𝜀b > 0, ∃ Nb(𝜀b) ∈ ℕ s.t. n ≥ Nb(𝜀b) ⇒ |an - B| < 𝜀b

So we know that ∀ 𝜀a > 0 and 𝜀b > 0, we can choose n large enough to make

|an - A| < 𝜀a and

|bn - B| < 𝜀b.

But we want to show

|(an + bn) - (A + B)| < 𝜀

What is the relation between |an - A|, |an - B|, and |(an + bn) - (A + B)|?
We do some algebra.

 |(an + bn) - (A + B)| = |an + bn - A - B| = |an - A + bn - B| = |(an - A) + (bn - B)| ≤ |an - A| + |bn - B| Using |X+Y| ≤ |X| + |Y| (triangle inequality - a property that needs to be proven)

Now we know that for n ≥ max(Na(𝜀a), Nb(𝜀b))

 |(an + bn) - (A + B)| ≤ |an - A| + |bn - B| < 𝜀a + |bn - B| < 𝜀a + 𝜀b

To make

|(an + bn) - (A + B)| < 𝜀

it would be sufficient to make

𝜀a + 𝜀b ≤ 𝜀.

Can we just pick any positive 𝜀a and 𝜀b so that 𝜀a + 𝜀b ≤ 𝜀?
Yes. Remember that the only requirements on 𝜀a and 𝜀b are that both are positive.
So ∀ 𝜀 > 0, we can choose 𝜀a = 𝜀b = 𝜀 / 2.
There is nothing magic about the factor of 1/2. We could choose other values (e.g. 𝜀a = 𝜀 / 3 and 𝜀b = 𝜀 / 100) as long as 𝜀a + 𝜀b ≤ 𝜀.

So we can define N(𝜀) = max(Na(𝜀a), Nb(𝜀b)).

Summary:

1. ∀ 𝜀 > 0, choose 𝜀a = 𝜀b = 𝜀 / 2
2. The convergence of an ⇒ Na(𝜀a) exists
3. The convergence of bn ⇒ Nb(𝜀b) exists
4. The existence of Na(𝜀a) and Nb(𝜀b) ⇒ N(𝜀) = max(Na(𝜀a), Nb(𝜀b)) exists
5. N(𝜀) exists and satisfies the conditions in the definition of convergence of (an + bn) to (A + B)
6. Therefore (an + bn) converges to (A + B)

## Writing out the Proof:

If an converges to A and bn converges to B,
then ∀ 𝜀 > 0, let 𝜀a = 𝜀b = 𝜀 / 2

 an converges to A ⇒ ∃ Na(𝜀a) ∈ ℕ s.t. n ≥ Na(𝜀a) ⇒ |an - A| < 𝜀a bn converges to B ⇒ ∃ Nb(𝜀b) ∈ ℕ s.t. n ≥ Nb(𝜀b) ⇒ |bn - B| < 𝜀b

max(Na(𝜀a), Nb(𝜀b)) exists.
Let N(𝜀) = max(Na(𝜀a), Nb(𝜀b)).
Then

 n ≥ N(𝜀) ⇒ |(an + bn) - (A + B)| = |an + bn - A - B| = |an + bn - A - B| = |an - A + bn - B| = |(an - A) + (bn - B)| ≤ |an - A| + |bn - B| Triangle Inequality: |X + Y| ≤ |X| + |Y| < 𝜀a + |bn - B| Since n ≥ Na(𝜀a) < 𝜀a + 𝜀b Since n ≥ Nb(𝜀b) = 𝜀/2 + 𝜀/2 By definition of 𝜀a and 𝜀b = 𝜀 n ≥ N(𝜀) ⇒ |(an + bn) - (A + B)| < 𝜀

If an converges to A and bn converges to B, then

∀ 𝜀 > 0, ∃ N(𝜀) ∈ ℕ s.t. n ≥ N(𝜀) ⇒ |(an + bn) - (A + B)| < 𝜀

(an + bn) converges to (A + B) by the defintion of convergence.

Key Takeaways:

1. Again, the order in which the proof is presented is not the order in which the proof is constructed.
Construction: Find 𝜀a and 𝜀b at the end.
Presentation: 𝜀a and 𝜀b are given at the beginning.
If one reads only the proof, key parts (like 𝜀a = 𝜀b = 𝜀 / 2 and N = max(Na(𝜀a), Nb(𝜀b))) seem to appear out of thin air.
The way to study the proof is to try the proof on your own and figure out the thought process that motivates the key parts.

2. This is an example of proving the existence of N without knowing what N is.

3. Patterns:
Much of the work in proofs is applying definitions and arithmentic, but there are some patterns that are repeated.
This example introduces a common pattern: splitting 𝜀.
Sometimes you want a sum of n terms to be less than 𝜀.
Check for conditions that force each term to be less than 𝜀/n.

4. Notation:
In cases with multiple converging sequences, adding a subscript or argument to the 𝜀 and N may reduce confusion (for the student and the grader).
In this example 𝜀a is the 𝜀 from the definition of convergence of an and Na(𝜀a) is the N associated with that 𝜀a.

Back to Real Analysis