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Example Proofs: Using Theorems
n! = n factorial
n! = 𝛱
_{k = 1 to n} k for n ≥ 1
So 1! = 1, 2! = 1*2, 3! = 1*2*3, ...
And 0! is defined to be 1.
Show that s
_{n} = ∑
_{i=0 to n}1/i! converges.
Thought Process:
Note: Using the definition of convergence
a_{n} converges to limit, L ⇔
∀ 𝜀 > 0, ∃ N ∈ ℕ s.t. n ≥ N ⇒ a_{n}  L < 𝜀
to prove the sequence converges requires knowing the limit.
Let's assume we don't recognize the series as e = ∑
_{i=0 to ∞}1/i!.
So we would look over the theorems on convergence.
Theorems have one or more conditions and some conclusion.
When using a theorem, you must prove that
every condition is met.
This can effectively lead to one proof per condition.
Here are two theorems that can be used to conclude that a sequence converges.

Cauchy Criterion:
A sequence a_{n} is said to be a Cauchy sequence if
∀ 𝜀 > 0, ∃ N ∈ ℕ s.t. n≥ N and m ≥ N ⇒ a_{n}  a_{m} < 𝜀
Theorem 1: A real valued sequence converges iff it is Cauchy.

Bounded Monotonic Sequences:
A sequence a_{n} is said to be monotonically increasing if a_{n} ≤ a_{n+1}
for n = 1, 2, 3, ...
A sequence a_{n} is said to be monotonically decreasing if a_{n} ≥ a_{n+1}
for n = 1, 2, 3, ...
Theorem 2: A monotonic sequence converges iff it is bounded.
So how do we decide which theorem to use?
In a way it doesn't matter. Neither theorem requires knowing the limit. Either one should work.
Theorem 1 has a single condition (the Cauchy property), but that condition is complicated.
Theorem 2 has two conditions (monotonic and bounded), but each condition is simple.
Let's try both.
Using Bounded Monotonicity (Theorem 2)
∀ integer n ≥ 1,



0


< 

1/(n+1)! 



s_{n} + 0 

< 

s_{n} + 1/(n+1)! 



s_{n}


< 

(∑_{i=0 to n} 1/i!) + 1/(n+1)! 





< 

∑_{i=0 to n+1} 1/i! 





< 

s_{n+1} 
So s
_{n} is monotonically increasing.
(intermediate result 1)
Is s
_{n} bounded?
The difficult part to work with is the factorials.
Trying to simplify the factorial, we note that if we want an upper bound on 1/i!,
we can look for a lower bound on i!.



1! 

= 

1 

= 

2^{0} 



2! 

= 

1*2 

= 

2^{1} 



3! 

= 

1*2*3 

> 

2^{2} 



4! 

= 

1*2*3*4 

> 

2^{3} 



i! 





≥ 

2^{i1} 


for i ≥ 1 



1/i! 





≤ 

1/2^{i1} 


for i ≥ 1 
Note: The presentation above illustrates the idea.
To be rigorous, you could use proof by induction to shiw that 1/i! ≤ 1/2
^{i1}.
Then



s_{n} 

= 

s_{n} 





= 

∑_{i=0 to n} 1/i! 





= 

1/0! + ∑_{i=1 to n} 1/i! 





= 

1/1 + ∑_{i=1 to n} 1/i! 





= 

1 + ∑_{i=1 to n} 1/i! 





≤ 

1 + ∑_{i=1 to n} 1/2^{i1} 





≤ 

1 + ∑_{i=0 to n1} (1/2)^{i} 



s_{n} 

≤ 

1 + ∑_{i=0 to n1} (1/2)^{i} 


(intermediate result 3)

Then we either remember or derive a closed form for b
_{n} = ∑
_{ i=0 to m} r
^{i},



(1r) * b_{n} 

= 


(1r) * ∑_{i=0 to m} r^{i} 





= 


∑_{i=0 to m} r^{i}  r * ∑_{i=0 to m} r^{i} 





= 

r^{0} 
+ r^{1} + r^{2} + ... + r^{n1} + r^{m} 








 r^{1}  r^{2}  ...  r^{m1}  r^{m} 
 r^{m+1} 





= 
1 


 r^{m+1} 



(1r) * b_{m} 

= 
1 


 r^{m+1} 



b_{m} 

= 


(1  r^{m+1}) / (1r) 



∑_{i=0 to m} r^{i} 

= 


(1  r^{m+1}) / (1r) 


(intermediate result 4) 
Then



s_{n} 

≤ 

1 + ∑_{i=0 to n1} (1/2)^{i} 



(intermediate result 3)






= 

1 + (1  (1/2)^{n}) / (11/2) 



using (intermediate result 4) with r = 1/2 and m = n1 





= 

1 + (1  (1/2)^{n}) / (1/2) 





= 

1 + 2  (1/2)^{n  1} 





< 

1 + 2  (1/2)^{n  1} + (1/2)^{n  1} 



since (1/2)^{n  1} > 0 





= 

1 + 2 + 0 







= 

3 





s_{n} 

< 

3 
So s
_{n} is bounded.
(intermediate result 2)
s
_{n} is monotonic and bounded by
(intermediate result 1) and
(intermediate result 2)
so s
_{n} converges by Theorem 2.
Notes:

Each property requires its own proof.

Because the properties were not too complicated, the work does not need to be reordered for the proof.

The actual use of the theorem is only a single sentence (2 lines).
Using The Cauchy Criterion (Theorem 1)
Assume n ≥ 0 and m ≥ 0. Without loss of generality, assume n ≤ m



s_{n}  s_{m} 

= 

∑_{i=0 to n}1/i!  ∑_{i=0 to m} 1/i! 


Definition of s_{n} 





= 

∑_{i=n+1 to m} 1/i! 





= 

∑_{i=n+1 to m} 1/i! 





< 

∑_{i=n+1 to m} 1/2^{i1} 


Same argument as in previous proof. 





= 

2 * ∑_{i=n+1 to m} 1/2^{i} 





< 

2 * (1/2)^{n + 1}∑_{i=0 to mn1} 1/2^{i} 





= 

(1/2)^{n} * (1  (1/2)^{m  n})/(1  1/2) 


Using (intermediate result 4) 





= 

(1/2)^{n} * (1  (1/2)^{m  n})/(1/2) 





= 

(1/2)^{n  1} * (1  (1/2)^{m  n}) 





< 

(1/2)^{n  1} * (1) 





= 

(1/2)^{n  1} 





= 

(1/2)^{n  1} 
The Cauchy property requires that a
_{n}  a
_{m} < 𝜀.
So we want to find values of n that satisfy (1/2)
^{n  1} < 𝜀.


(1/2)^{n  1} < 𝜀 


⇔ 


2*(1/2)^{n} < 𝜀 





⇔ 


(1/2)^{n} < 𝜀/2 





⇔ 


2^{n} > 2/𝜀 





⇔ 


n > log_{2}(2/𝜀) 
Writing out the Proof:
Let s
_{n} = ∑
_{i=0 to n}1/i!
∀ 𝜀 > 0, let N = ceiling(log
_{2}(2/𝜀))
Then let n and m be integers s.t. n≥N and m≥N. (without loss of generality, assume n ≤ m.)



s_{n}  s_{m} 

= 

∑_{i=0 to n} 1/i!  ∑_{i=0 to m} 1/i! 


Definition of s_{n} 





= 

∑_{i=n+1 to m} 1/i! 





= 

∑_{i=n+1 to m} 1/i! 





< 

∑_{i=n+1 to m} 1/2^{i1} 


Same argument as in previous proof. 





= 

2 * ∑_{i=n+1 to m} 1/2^{i} 





< 

2 * (1/2)^{n + 1}∑_{i=0 to mn1} 1/2^{i} 





= 

(1/2)^{n} * (1  (1/2)^{m  n})/(1  1/2) 


Using (intermediate result 4) 





= 

(1/2)^{n} * (1  (1/2)^{m  n})/(1/2) 





= 

(1/2)^{n  1} * (1  (1/2)^{m  n}) 





< 

(1/2)^{n  1} * (1) 





= 

(1/2)^{n  1} 





= 

2 * (1/2)^{n} 





≤ 

2 * (1/2)^{N} 


since n ≥ N 





≤ 

2*(1/2)^{log2(2/𝜀)} 


By choice of N 





= 

2*1/2^{log2(2/𝜀)} 





= 

2*1/(2/𝜀) 





= 

𝜀 



s_{n}  s_{m} 

< 

𝜀 
∀ 𝜀 > 0, ∃ N = ceiling(log
_{2}(2/𝜀)) ∈ ℕ
s.t. for n,m ∈ ℕ, n≥N and m≥N ⇒ s
_{n}  s
_{m} < 𝜀
s
_{n} = ∑
_{i=0 to n}1/i! is a Cauchy sequence.
s
_{n} converges by Theorem 1.
Key Takeaways:

Again, the order in which the proof is presented is not the order in which the proof is constructed.
Construction: N is found at the end.
Presentation: N is given at the beginning.
If one reads only the proof, the key part (N = ceiling(log_{2}(2/𝜀))) seems 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 page shows an example of how you can prove a single result in multiple ways.

Patterns:
When applying theorems, you must make sure that every condition is met.
The proof will be a sequence of subproofs, one for each property.
After those, applying the theorem may be just one line.

Notation:
When your proofs get long, you might want to label intermediate results for easier reference later in the proof.
Using different colors or boxes around key points makes your proof easier to follow.

Organizing the theorems:
This example considered only two theorems.
But a typical class will have many theorems.
One way to organize the theorems in you head is to group them by inputs and outputs.
Inputs are the conditions that must be met (monotonic, bounded, Cauchy).
Outputs are the conclusions (the squence converges, the sequence converges to limit L,
the function is differentiable).
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