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

Using the Definition of Convergence

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

show that if a_n converges to limit, A and b_n converges to limit, B,
then a_n + b_n 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 ⇒ |(a_n + b_n) - (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 ∀ 𝜀_a > 0 and 𝜀_b > 0, we can choose n large enough to make

|a_n - A| < 𝜀_a and

|b_n - B| < 𝜀_b.

But we want to show

|(a_n + b_n) - (A + B)| < 𝜀

What is the relation between |a_n - A|, |a_n - B|, and |(a_n + b_n) - (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(N_a(𝜀_a), N_b(𝜀_b))

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

To make

|(a_n + b_n) - (A + B)| < 𝜀

it would be sufficient to make

𝜀_a + 𝜀_b ≤ 𝜀.

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

So we can define N(𝜀) = max(N_a(𝜀_a), N_b(𝜀_b)).

Summary:

1. ∀ 𝜀 > 0, choose 𝜀_a = 𝜀_b = 𝜀 / 2
2. The convergence of a_n ⇒ N_a(𝜀_a) exists
3. The convergence of b_n ⇒ N_b(𝜀_b) exists
4. The existence of N_a(𝜀_a) and N_b(𝜀_b) ⇒ N(𝜀) = max(N_a(𝜀_a), N_b(𝜀_b)) exists
5. N(𝜀) exists and satisfies the conditions in the definition of convergence of (a_n + b_n) to (A + B)
6. Therefore (a_n + b_n) converges to (A + B)

Writing out the Proof:

If a_n converges to A and b_n converges to B,
then ∀ 𝜀 > 0, let 𝜀_a = 𝜀_b = 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(N_a(𝜀_a), N_b(𝜀_b)) exists.
Let N(𝜀) = max(N_a(𝜀_a), N_b(𝜀_b)).
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 a_n converges to A and b_n converges to B, then

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

(a_n + b_n) 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 𝜀_a and 𝜀_b at the end.
   Presentation: 𝜀_a and 𝜀_b are given at the beginning.
   If one reads only the proof, key parts (like 𝜀_a = 𝜀_b = 𝜀 / 2 and N = max(N_a(𝜀_a), N_b(𝜀_b))) 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 𝜀_a is the 𝜀 from the definition of convergence of a_n and N_a(𝜀_a) is the N associated with that 𝜀_a.

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