Subject: Calculus

Limit Of A Sequence

Limits of a Sequence at a Glance

What is a Sequence?

Well, in layman's terms, it is a set of things that are in order. In mathematics, it is a set of numbers that are in order (I know I know, the difference is so obvious). You can think of things like the numbers from 1 to 10 as a sequence (you know: 1, 2, 3, 4... do I really have to go all the way here... 9, 10) and many more. Basically, there are two types of mathematical sequences: the finite and the infinite.

Finite Sequence and Its Limit

Finite sequences are those ordered set of numbers that have a last term. For example:

3, 6, 9, 12, 15, 18, 21

is a finite sequence. There are two limits to a finite sequence. The higher bound limit and the lower bound limit of a finite sequence correspond to the maximum and minimum values of the sequence respectively. So, in the case of the above example, the higher bound or upper bound limit is 21 while its lower bound limit is 3. From here, let's focus on the limits of an infinite sequence.

Infinite Sequence and Its Limit

An infinite sequence are ordered set of numbers that go on forever. An example will clear things up:

\frac{1}{3}, \frac{2}{5},\frac{3}{7},\frac{4}{9},\frac{5}{11},...

The three dots (AKA: an ellipsis for you smart students out there) that appear at the end simply tells us that set of ordered numbers continues on infinitely. Infinite sequences are best represented by a function. Without this, it would take forever to define an infinite sequence (hence the name). A sequence function is a function whose domain is the set of all positive integers. For example, the infinite sequence presented above is defined by the following sequence function:

f(n)=\frac{n}{2n+1}

where n assumes any values from 1 to infinity. When n=1 we get f(1)=\frac{1}{3}, when n=2, f(2)=\frac{2}{5} and so on as in the above example. But where does it all end? Or does this sequence end? Remember as a child when you asked your parents "how long till we get there?" and they would say, "20 minutes" but it always seemed to be 20 hours? In order to see if the sequence ends, we should take n to a very large number and see what happens to the sequence. Note that as n goes very large, the value in the denominator of the function, 2n+1 goes to 2n (if n = 1,000,000 then an extra 1 on top of that won't make much of a difference). With that little piece of information, we can simplify the function to:

f(n)=\frac{n}{2n+1}\simeq \frac{n}{2n}=\frac{1}{2}

So as the sequence continues, it will approximately end up with the value of \frac{1}{2}. That value is called the limit of the sequence function f.

Generally, L is the limit of a sequence a_{n}, where f(n)=a_{n}. For a very large number n, |a_{n}-L| is very small (similar to the steps for a derivative). This also explains why the 1 in the above example could be ignored (for a large n it wasn't important). The definition of the limit of an infinite sequence is basically the same to the definition of the limit of any ordinary function. So in evaluating limits of a sequence, we may treat it as any ordinary function and then apply the limit theorems that you should know by now.

Example Problems

Example #1

Find the limit of the following sequence function.

f(n)=\frac{1}{n}

Solution: As discussed above, we can take the sequence function just like any ordinary function in evaluating its limits. Thus,

\lim_{n \to \infty} f(n)) = \lim_{x \to \infty}f(x)
 = \lim_{x \to \infty}\frac{1}{x}
 =0

Thus, the limit of the sequence a_{n}=\frac{1}{n}=1, \frac{1}{2}, \frac{1}{3},...,\frac{1}{n} is zero.

Example #2

Find the limit of the following sequence function.

a_{n}=\frac{4n^{2}}{2n^{2}+1}

Solution:To solve this one, we first transform a_{n} to f(x) to obtain

f(x)=\frac{4x^{2}}{2x^{2}+1}

then we solve the limit by dividing both the numerator and the denominator by x^{2}. Thus

\lim_{x \to \infty} f(x)=\lim_{x \to \infty}\frac{4x^{2}}{2x^{2}+1}
 =\lim_{x \to \infty}\frac{\frac{4x^{2}}{x^{2}}}{\frac{2x^{2}}{x^{2}}+\frac{1}{x^{2}}}
 =\lim_{x \to \infty}\frac{4}{2+\frac{1}{x^{2}}}
 =\lim_{x \to \infty}\frac{4}{2+0}
 =2

Thus, the limit of the sequence function a_{n} is 2.

Example #3

Find the limit of the following sequence function.

a_{n}=n

The limit of the above sequence Does not exist

See the note below for more details


Convergent & Divergent Sequences

Obviously, the sequence a_{n}=n blows up as n goes to infinity. Therefore, the limit of a_{n} as n goes to infinity does not exist. This kind of sequences are called divergent. They are sequences whose limits are undefined or infinite. The other classification of sequences are those convergent. Convergent sequence are those whose limits do exist and its point of convergence is the value of the limit itself. Examples 1 and 2 above are best representation of a convergent sequence.

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