We say that is the limit of the sequence as goes to infinity if , then If such an exists, we say that the sequence is convergent and write We may also use the notation to mean converges to . If no such exists, then we say that the sequence diverges.

We will use this formal definition to show that . We will take a sequence where .

Due to the Archimedean Property. No matter what we are given, we can always find a cutoff so that . An example of we can take is .

We say that if for every tolerance , the interval contains a Tail of the sequence .

The following statements can all be viewed as being equivalent:

2. Every interval contains a Tail of .

3. Every interval contains all but finitely many terms of .

4. Every interval containing contains a Tail of .
   

5. Every interval containing contains all but finitely many terms of .

Important Note: Changing finitely many terms in does not affect convergence, i.e. every Subsequence of a sequence converges to iff the sequence converges to

Let be limits of .
For Contradiction, we assume
WLOG
Let

We can construct two non-overlapping intervals, each containing one of the limits. Specifically, use the intervals and .

Therefore both these intervals must contain the Tail of the sequence. This contradicts

Therefore

or does not imply the existence of a limit. The notation simply gives us a way of describing the behavior of the sequence far out in the tail.

Proof

Proof
We have two cases and

Case is trivial because of Property 1 (Anything multiplied by 0 is 0)

Case

Let . Then since we know that converges to , we can find a cutoff such that if , then

It follows that if , then

Applying the Triangle Inequality here
So let's suppose that we are given a tolerance . If we can make both and , then would tell us that But since , we can find a cutoff so that if , then Similarly, since , we can find a cutoff so that if , then Now we let , the largest of the two cutoffs. Then if , we have that and , simultaneously. This is the required cutoff.

Observe that

Let

Proof Real Analysis Theorems > Theorem 1

If we can prove that it is sufficient to show Property 4 is true due to Property 4 provided that . If then the limit may or may not exist Real Analysis Theorems > Theorem 3

Using Triangle Inequality

Real Analysis Theorems > Theorem 2
Since ,
Case
For our property to be true, the Since
Let . Since , we can find a cutoff such that if , then

Then

Case
Rearranging Series > Geometric Series.

and hence (if )

If
Let and . We get

Since

If then

Therefore when
Using Property 4, we can take and prove that

Using this we have proven for 𝟘, Using Property 5, taking sequence in numerator to be 1 for all n and sequence in denominator to be , we can prove for .
#TA

This rule is really just an observation that convergence is about the behaviour of the tail of a sequence. We can in fact change any finite number of the terms in a sequence without impacting convergences.

We use to help find the limit of recursively defined sequences. This has been used in HW Assignments Crowdmark.pdf > page=16