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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:

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.

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. ..."



a_{n} converges to A 


⇒ 


∀ 𝜀_{a} > 0, ∃ N_{a}(𝜀_{a}) ∈ ℕ s.t. n ≥ N_{a}(𝜀_{a}) ⇒ a_{n}  A < 𝜀_{a} 



b_{n} converges to B 


⇒ 


∀ 𝜀_{b} > 0, ∃ N_{b}(𝜀_{b}) ∈ ℕ s.t. n ≥ N_{b}(𝜀_{b}) ⇒ a_{n}  B < 𝜀_{b} 
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.



(a_{n} + b_{n})  (A + B) 


= 


a_{n} + b_{n}  A  B 






= 


a_{n}  A + b_{n}  B 






= 


(a_{n}  A) + (b_{n}  B) 






≤ 


a_{n}  A + b_{n}  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}))



(a_{n} + b_{n})  (A + B) 


≤ 


a_{n}  A + b_{n}  B 






< 


𝜀_{a} + b_{n}  B 






< 


𝜀_{a} + 𝜀_{b} 
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:
 ∀ 𝜀 > 0, choose 𝜀_{a} = 𝜀_{b} = 𝜀 / 2
 The convergence of a_{n} ⇒ N_{a}(𝜀_{a}) exists
 The convergence of b_{n} ⇒ N_{b}(𝜀_{b}) exists
 The existence of N_{a}(𝜀_{a}) and N_{b}(𝜀_{b}) ⇒ N(𝜀) = max(N_{a}(𝜀_{a}), N_{b}(𝜀_{b})) exists
 N(𝜀) exists and satisfies the conditions in the definition of convergence of (a_{n} + b_{n}) to (A + B)
 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



a_{n} converges to A^{ } 


⇒ 


∃ N_{a}(𝜀_{a}) ∈ ℕ s.t. n ≥ N_{a}(𝜀_{a}) ⇒ a_{n}  A < 𝜀_{a} 



b_{n} converges to B^{ } 


⇒ 


∃ N_{b}(𝜀_{b}) ∈ ℕ s.t. n ≥ N_{b}(𝜀_{b}) ⇒ b_{n}  B < 𝜀_{b} 
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) 


⇒ 


(a_{n} + b_{n})  (A + B) 

= 

a_{n} + b_{n}  A  B 











= 

a_{n} + b_{n}  A  B 











= 

a_{n}  A + b_{n}  B 











= 

(a_{n}  A) + (b_{n}  B) 











≤ 

a_{n}  A + b_{n}  B 


Triangle Inequality: X + Y ≤ X + Y 











< 

𝜀_{a} + b_{n}  B 


Since n ≥ N_{a}(𝜀_{a}) 











< 

𝜀_{a} + 𝜀_{b} 


Since n ≥ N_{b}(𝜀_{b}) 











= 

𝜀/2 + 𝜀/2 


By definition of 𝜀_{a} and 𝜀_{b} 











= 

𝜀 



n ≥ N(𝜀) 


⇒ 


(a_{n} + b_{n})  (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:

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.

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

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.

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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