Knowledge Node - Sequence Limits

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In Calculus we are introduced to the limit process without a truly formal definition of its meaning or inner working. In fact, most of Calculus is presented in an intuitive manner.

Suppose we are given the sequence (an), and we are told the following:

(an) = a

This essentially means that the sequence converges to the given value, a. A different way to write the same idea is that

(an) a.

What does it mean to converge? We know intuitively that an example sequence, such as , converges to 0, because as n gets much much larger, the entire fraction moves towards 0. But how can we prove this? What if the convergence isn't so obvious?

The formal expression of the limit has undergone many changes over the last several hundred years. One of it's present forms is as follows:

Definition. For an arbitrary sequence (an), the sequence is said to converge to some real number L if the condition below is met.

Given any > 0, there exists an N such that for all n  that satisfy n > N, the following is true:

| an - L | <

What this essentially means is that no matter what arbitrarily small is chosen, we can find a cut off point where the distance between (the limit) and (every number in the sequence after that cutoff) is less than that .

Ex. Suppose we wish to prove that the example sequence above really does converge to 0.

Given that we wish to prove that

an = 0, or that (an) 0.

First, let's do a little scratch work.

By our formal definition of convergence, we know that . Since L = 0, and n is a positive whole number (a natural number) we can simplify this inequality to . Since we want to find an expression so that all n are greater than , we multiply both sides by the denominator and divide both sides by to get . Subtracting 2 from both sides yields . We can divide both sides by 3 and get . Finally, if we take the square root of both sides, we have .

Now we can begin our proof.

Proof. Let any > 0 be given. Let N such that . Likewise, we let n such that n > N. Therefore, .

Since we have this relationship, we also have . With some relatively simple modifications we can manipulate our equation back into a form that clearly shows that our sequence is always less than , and therefore our distance from 0, our limit, is less than .

First, we square both sides, . Then we multiply by three and get . Next we add two for , we divide by the right side and multiply by and get , which shows that our sequence value at any n, when n is chosen to be greater than or equal to the specific N above, is within our desired distance. Similar proofs work for nearly ever sequence.

Definition. A sequence is said to diverge if it does not converge to any value. A definition can be constructed using the negation of the convergence definition. Essentially, we must find an value such that no matter what N we pick, there exists an n > N, such that we have | an - L | > . In other words, no matter what cut off point we pick, there is at least one n in the sequence that doesn't fit the criteria.

just curious... added 5/4/04

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