Limit of a Sequence

Go back to the sequence \( \{s_n \} = \{e^{-n} +1 \} \) . The candidate for the limit is \( L = 1 \). Graph the points of the sequence along with the candidate limit \( L = 1 \) , and draw in the lines \( y = L + \epsilon = 1 + \epsilon \) and \( y = L - \epsilon = 1 - \epsilon \).

Fig. 1. Trapping the sequence values.

You can see that no matter how tiny \( \epsilon \) is, you will always be able to go out far enough (in other words, pick out a big enough \( M \) ) so that the sequence values are trapped between the lines \( y = 1 + \epsilon \) and \( y = 1 + \epsilon \). That means the sequence converges to the limit \( L = 1 \).

Going back to the sequence \( \{s_n \} = \{e^{-n} +1 \} \) , use the properties of limits for sequences to find the limit as \( n \to \infty \) .

Answer:

Using the Sum Rule,

\[ \begin{align} \lim\limits_{n \to \infty} s_n & = \lim\limits_{n \to \infty} (e^{-n} +1 ) \\ &= \lim\limits_{n \to \infty} e^{-n} +\lim\limits_{n \to \infty} 1 \\ &= 0 + 1 \\ &= 1. \end{align} \]

Example 1: Take the sequences \( \{ s_n \} = \{ n \} \) and

\[ \{ t_n \} = \left\{ \frac{1}{n} \right\}. \]

Then \( \{ s_n \} \) diverges while \( \{ t_n \} \to \infty \). But

\[ \begin{align} \lim\limits_{n \to \infty} (s_n \cdot t_n ) &= \lim\limits_{n \to \infty} n \cdot \frac{1}{n} \\ &= \lim\limits_{n \to \infty} 1 \\ &= 1 . \end{align} \]

So here you get 1 for the limit of the product.

Example 2: Can you get something else for the limit of the product if the limit of one of the sequences doesn't exit? If you instead take the sequence

\[ \{ w_n \} = \left\{ \frac{1}{n^2} \right\}, \]

then \( \{ w_n \} \to 0 \) while

\[ \begin{align} \lim\limits_{n \to \infty} (s_n \cdot w_n ) &= \lim\limits_{n \to \infty} n \cdot \frac{1}{n^2} \\ &= \lim\limits_{n \to \infty} \frac{1}{n} \\ &= 0 . \end{align} \]

Here you got 0 for the limit of the product, which is definitely not the same as what you got in the first example.

Example 3: Can you arrange for the limit of the product to diverge if the limit of one of the sequences is zero, but the limit of the other sequence doesn't exist? What if \( \{ z_n \} = \{ n^2 \} \) ? Then

\[ \begin{align} \lim\limits_{n \to \infty} (z_n \cdot t_n ) &= \lim\limits_{n \to \infty} n^2 \cdot \frac{1}{n} \\ &= \lim\limits_{n \to \infty} n , \end{align} \]

and the product diverges. So, you can get the limit of the product not existing!

So if you don't have the conditions correct to use the Product Rule, anything can happen, and you can't predict what it might be in advance!

Does the sequence

\[ \{ s_n \} = \left\{ 2 + \frac{4}{n} \right\} \]

have a limit? If so, what is it?

Answer:

Another way of framing this question is, "does the above sequence approach a single value as \( n \) gets large? Let's see!

In the question, there is a \( \frac{4}{n} \) term. Let's look at the function equivalent of this. For the function

\[ f(x) = \frac{1}{x} \]

you know that

\[ \begin{align} \lim\limits_{x \to \infty} f(x) &= \lim\limits_{x \to \infty} \frac{1}{x} \\ &= 0 \end{align} \]

because the function has a horizontal asymptote of \( y =0 \). That means the sequence

\[ \{ t_n \} = \left\{ \frac{1}{n} \right\} \]

also has

\[ \begin{align} \lim\limits_{n \to \infty} t_n &= \lim\limits_{n \to \infty} \frac{1}{n} \\ &= 0 \end{align} \]

since the sequence is the same as the function except for the domain. In fact, you can see it graphically as well.

Fig. 2. Graph of the sequence {1/n} on the positive x-axis.

Now that we've reminded ourselves of the characteristics of a reciprocal function, let's get back to the original question. Now you know you can apply the Sum Rule to get

\[ \begin{align} \lim\limits_{n \to \infty} s_n &= \lim\limits_{n \to \infty} \left( 2 + \frac{4}{n} \right) \\ &= \lim\limits_{n \to \infty} 2 + \lim\limits_{n \to \infty} \frac{4}{n}, \end{align} \]

and then the Constant Rule to get:

\[ \begin{align} \lim\limits_{n \to \infty} 2 + \lim\limits_{n \to \infty} \frac{4}{n} &= 2 + 4 \lim\limits_{n \to \infty} \frac{1}{n} \\ &= 2 + 4 \cdot 0 \\ &= 2. \end{align} \]

So the sequence does have a limit, and the value is 2.

Does the sequence

\[ \left\{ \frac{1 + 4n}{5 + 6n} \right\} \]

Answer:

Sometimes you will need to try different things to find the one that lets you use the rules correctly. You would like to use the Quotient Rule to solve this problem.

First try setting up two sequences, \( \{ s_n \} = \{ 1 + 4n \} \) and \( \{ t_n \} = \{ 5 + 6n \} \). Oops, there is a problem since the Quotient Rule requires both of those sequences to have a limit, and neither one converges to a finite number!

For the second try, break it up into two fractions instead of just one. You know that

\[ \frac{1+4n}{5+6n} = \frac{1}{5+6n} + 4 \cdot \frac{n}{5 + 6n}, \]

which is definitely closer to being useful, but still not quite there because of that

\[ \frac{n}{5+6n} \]

term.

The second try gives you the idea that you will want to factor an \( n \) out of the denominator first. Then you have

\[ \frac{1+4n}{5+6n} = \frac{1+4n}{n \left( \frac{5}{n}+6 \right) } . \]

Algebra to the rescue! Now set up the two sequences to use the Quotient Rule,

\[ \{ s_n \} = \left\{\frac{1}{n}+4 \right\} \mbox{ and } \{ t_n \} = \left\{ \frac{5}{n} + 6 \right\}. \]

Both of those have limits, in fact

\[ \lim\limits_{n \to \infty} s_n = \lim\limits_{n \to \infty} \left( \frac{1}{n}+4 \right) = 4 \]

and

\[ \lim\limits_{n \to \infty} t_n = \lim\limits_{n \to \infty} \left( \frac{5}{n}+6 \right) = 6 \]

where you have applied the Sum Rule and the Constant Rule as in the previous example. Now you know you can apply the Quotient Rule to get

\[ \begin{align} \lim\limits_{n \to \infty} \frac{1 + 4n}{5 + 6n} &= \lim\limits_{n \to \infty} \frac{s_n}{t_n} \\ &= \frac{4}{6} \\ &= \frac{2}{3}. \end{align} \]

Therefore the sequence does converge, and the limit is \( \frac{2}{3} \).

Does the sequence \( \{ s_n \} = \left\{ (-1)^n \right\} \) converge? If so, what does it converge to?

Answer:

To get an idea of how this sequence behaves, let's write out some of the terms of this sequence.

\[ \{-1, 1, -1, 1, \dots \} \]

If the sequence has a limit, the limit would need to be either \( -1 \) or \( 1 \) since those are the only two values in the sequence and they don't change at all. Let's see what happens graphically when you try to choose \( L = 1 \) for the limit value.

Fig. 3. Is L=1 the limit for the sequence?

You can see looking at the picture above that it doesn't matter how large an \( M \) you pick, there is no way to get all of the sequence values to be between the two lines \( y = 1 + \epsilon \) and \( y = 1 - \epsilon \). That means this sequence doesn't converge.

You can also say that the sequence diverges.

Let's look at the sequence

\[ \{ s_n \} = \left\{ \frac{ (-1)^n}{n} \right\}. \]

Answer:

Using the Absolute Value Theorem,

\[ \begin{align} \lim\limits_{n \to \infty} \left| s_n \right| &= \lim\limits_{n \to \infty} \left| \frac{ (-1)^n}{n} \right| \\ &= \lim\limits_{n \to \infty} \frac{ 1}{n} \\ &= 0, \end{align} \]

so

\[ \lim\limits_{n \to \infty} \frac{ (-1)^n}{n} =0 \]

as well.

Why is it important that the limit of the sequence in the Absolute Value Theorem is zero?

Answer:

Take the sequence \( \{ s_n \} = \left\{ (-1)^n \right\} \) . From the work you did above, you know this sequence doesn't converge, but

\[ \begin{align} \lim\limits_{n \to \infty} \left| s_n \right| &= \lim\limits_{n \to \infty} \left| (-1)^n \right| \\ &= \lim\limits_{n \to \infty} 1 \\ &= 1. \end{align} \]

So even though the absolute value of the sequence converges, the sequence itself does not. So verifying the condition that the limit of the absolute value of the sequence is zero by applying the Absolute Value Theorem is very important!

Consider the sequence \( \{ s_n \} = \left\{ 2^n \right\} \). Since

\[ \lim\limits_{n \to \infty} s_n =\lim\limits_{n \to \infty} 2^n = \infty , \]

the sequence \( \{ s_n \} \) diverges to infinity.