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Limit of a Sequence

- Calculus
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Think about the function \( f(x) = e^{-x}+1 \) and take the limit as \( x \to \infty \). You don't care about function values when \( x \) is small. You only care about function values as \( x \) gets very large because you are taking a "limit as \( x \) goes to infinity".

The same applies to sequences. Take the sequence \( \{s_n \} = \{e^{-n} +1 \} \). This is exactly the same as the function above, except the domain is now the natural numbers instead of the real numbers. If you want to know the "limit as \( n \) goes to infinity," you will be looking at very large values of \( n \), just like you looked for very large values of \( x \).

So keeping in mind that the process will be very similar to looking at the limits of sequences and functions, let's dive in!

See Limits of a Function for a review of functions and how to take their limits.

First, let's take a look at an informal definition of a limit of a sequence:

The **limit of a sequence** is the value the sequence approaches as the number of terms gets very large.

More formally:

Let \( L \) be a real number. The sequence has the limit \( L \) as \( n \) approaches \( \infty \) if given \( \epsilon > 0 \) , there exists a number \( M > 0 \) such that \( n > M \) implies \( \left| s_n - L \right| < \epsilon \). We write that

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

and say that the sequence **converges** to \( L \) . Sequences that do not have a limit are said to **diverge**.

Taking the limit of a function as \( x \to \infty \) you took a candidate for the limit (call it \( L \) for convenience), and then checked to see if you could "trap" the function values close to \( L \) as long as \( x \) was big enough.

Before going any further, let's look at a picture of what happens.

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 \( y = L - \epsilon = 1 - \epsilon \) .

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 \( L = 1 \).

There are two main ways to write "the limit of the sequence as \( n \) goes to infinity equals \( L \)", and you can use either of them:

\[ \{ s_n \} \to L; \] or

\[ \lim\limits_{n \to \infty} s_n = L . \]

Both mean the same thing. You can also say the sequence \( \{s _n \} \) converges to \( L \).

Naturally, you don't want to pick a candidate for the limit and then have to find an appropriate \( M \) that is big enough every time you want to show a sequence converges and what it converges to. Thankfully because sequences are functions, you can use the same limit rules for functions as you do for sequences.

Before talking about the uniqueness of a limit of a sequence, let's think about the solution of a linear equation. We say the linear equation \[ ax+b=0, \] where \( a \) and \( b \) are real numbers, has a unique solution. This means only one \( x \) value satisfies any given pair of values \( a \) and \( b \).

We can say the same about the limit of a sequence. If a sequence converges to a value and therefore has a limit, we say that this limit is unique to that sequence.

Suppose you have two sequences \( \{s _n \} \) and \( \{t _n \} \) , and you know that both converge. In other words, there exists numbers \( L \) and \( P \) such that

\[ \lim\limits_{n \to \infty} s_n = L \mbox{ and } \lim\limits_{n \to \infty} t_n = P . \]

Then the following rules hold:

**Sum Rule:**

\[ \lim\limits_{n \to \infty} (s_n + t_n ) = \lim\limits_{n \to \infty} s_n + \lim\limits_{n \to \infty} t_n = L + P . \]

**Difference Rule:**

\[ \lim\limits_{n \to \infty} (s_n - t_n ) = \lim\limits_{n \to \infty} s_n - \lim\limits_{n \to \infty} t_n = L - P . \]

**Product Rule:**

\[ \lim\limits_{n \to \infty} (s_n \cdot t_n ) = \left( \lim\limits_{n \to \infty} s_n \right) \cdot \left( \lim\limits_{n \to \infty} t_n \right) = L \cdot P . \]

**Constant Multiple Rule:**for any constant \( C \),

\[ \lim\limits_{n \to \infty} (C \cdot s_n ) = C\cdot \lim\limits_{n \to \infty} s_n = C \cdot L. \]

**Quotient Rule:** If \( P \not= 0 \) and \( t_n \not= 0 \) for all \( n \in \mathbb{n} \), then

\[ \lim\limits_{n \to \infty} \left( \frac{s_n}{t_n} \right) = \frac{\lim\limits_{n \to \infty} s_n }{ \lim\limits_{n \to \infty} t_n }= \frac{L}{P} . \]

It must be known both limits you are working with will converge for these properties to hold true!

So how do the properties of limits of sequences help you understand that if a sequence converges, the limit has to be unique?

Well, suppose you had a sequence that converged to two different things, say \( \{ s_n \} \to L\) and \( \{ s_n \} \to P\) , with \( L \not= P \). Then you can use the Difference Rule to say that

\[ \lim\limits_{n \to \infty} (s_n - s_n ) = \lim\limits_{n \to \infty} s_n - \lim\limits_{n \to \infty} s_n = L - P . \]

But wait a minute, \( s_n - s_n = 0 \), so it is also true that

\[ \lim\limits_{n \to \infty} (s_n - s_n ) = \lim\limits_{n \to \infty} 0 = 0.\]

Now you know that \( L - P = 0 \), or in other words, that \( L = P \). So really, you didn't have two different limits after all!

This is called a "proof by contradiction" and is a common math technique. First, you assume something, then show what you assumed actually couldn't have been true to start with.

Let's practice using some of these properties we just looked at!

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,

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

Making sure the conditions to use the rules for sequences are met is very important. Remember that you must know that both sequences converge and that if you use the quotient rule, the one in the denominator has a nonzero limit. If those aren't true, anything can happen!

What happens if one of your sequences doesn't converge? Even if the limit of the product exists, you can't do multiplication with something that doesn't exist. The following three examples will show you what can happen if both limits don't converge.

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!

Let's look at more examples of what kinds of limits a function can have and cases where it doesn't have a limit.

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.

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

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

converge? If so, what does it converge to?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

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

It would be very nice to cancel out that \( n \) in the denominator with one in the numerator, but to do that, you will need to factor it out first: \[ \begin{align} \frac{1+4n}{5+6n} & =\frac{n \left(\frac{1}{n}+4 \right) }{n \left( \frac{5}{n}+6 \right) } \\ &= \frac{ \frac{1}{n} + 4}{ \frac{5}{n} + 6}. \end{align} \]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 \( \frac{2}{3} \).

You can make this problem shorter by remembering the properties of rational functions. If the highest power in the numerator is the same as the highest power in the denominator, you can "divide" the coefficients to get the limit. In this case, the highest power in the numerator is *4n*, and the highest power in the denominator is *6n*, so dividing gives 4/6 = 2/3, which is both the limit and tells you that *y = 2/3* is the equation of the horizontal asymptote.

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.

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**.

Sometimes you will come up against a sequence like

\[ \left\{ \frac{ \cos n }{n} \right\} \]

where the Properties of Limits for Sequences can't be applied. In a case like this, the Squeeze Theorem can be helpful. Because sequences are just a special kind of function, the Squeeze Theorem can be restated for sequences.

To review the Squeeze Theorem for functions, see The Squeeze Theorem .

**Squeeze Theorem:** Suppose that there are two sequences \( \{ s_n \} \) and \( \{ t_n \} \), both of which converge to the same value \( L \), and that there exists an \( N \ in \mathbb{N} \) such that \( s_n \ le w_n \le t_n \) for all \( n \ge N \). Then

\[ \lim\limits_{n \to \infty} w_n = L . \]

Let's see how the Squeeze Theorem is applied. Going back to the sequence

\[ \left\{ \frac{ \cos n }{n} \right\}, \]

the idea is to "squeeze" it between two sequences that you know converge. First, let's look at a graph of some of the values of this sequence.

It certainly looks like it converges to zero, but you need to find the two sequences that you know converge to zero to "squeeze" it between. One sequence you have already worked with that converges to zero is the sequence

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

You also know that \( -1 \le \cos n \le 1 \) for any \( n\), so

\[ - \frac{1}{n} \le \frac{ \cos n}{n} \le \frac{1}{n} \]

for any \( n \) as well. That means you can take the second sequence you need to squeeze with to be

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

Taking a look at the graph for all three sequences,

So using the Squeeze Theorem for Sequences proves that the sequence

\[ \{ w_n \} = \left\{ \frac{ \cos n }{n} \right\} \]

converges.

There is a very handy consequence of the Squeeze Theorem for Sequences called the Absolute Value Theorem.

**Absolute Value Theorem:** Let \( \{ s_n \} \) be a sequence. If

\[ \lim\limits_{n \to \infty} \left| s_n \right| = 0, \]

then

\[ \lim\limits_{n \to \infty} s_n = 0. \]

You absolutely positively must know that the limit of the absolute value of the sequence is zero to apply this theorem!

Let's look at the sequence

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

Does it converge? If so, what does it converge to?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!

Sometimes a sequence just keeps growing larger and larger, like with the sequence. That is a somewhat nicer situation than one that just keeps jumping around, but it still doesn't converge. Instead, it has a special name.

If the sequence \( \{ s_n \} \) is such that

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

then we say that the sequence diverges to \( \pm \infty \).

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.

- Let \( L \) be a real number. The sequence has the limit \( L \) as \( n \) approaches \( \infty \) if given \( \epsilon > 0 \) , there exists a number \( M > 0 \) such that \( n > M \) implies \( \left| s_n - L \right| < \epsilon \). We write that
\[ \lim\limits_{n \to \infty} s_n = L, \]

and say that the sequence

**converges**to \( L \) . Sequences which do not have a limit are said to**diverge**. If the sequence \( \{ s_n \} \) is such that

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

then we say that the sequence diverges to \( \pm \infty \).

**Squeeze Theorem:**Suppose that there are two sequences \( \{ s_n \} \) and \( \{ t_n \} \), both of which converge to the same value \( L \), and that there exists an \( N \ in \mathbb{N} \) such that \( s_n \ le w_n \le t_n \) for all \( n \ge N \). Then\[ \lim\limits_{n \to \infty} w_n = L . \]

**Absolute Value Theorem:**Let \( \{ s_n \} \) be a sequence. If\[ \lim\limits_{n \to \infty} \left| s_n \right| = 0, \]

then

\[ \lim\limits_{n \to \infty} s_n = 0. \]

If a sequence converges, it has a unique limit.

Suppose you have two sequences \( \{s _n \} \) and \( \{s _n \} \) , and there exists numbers \( L \) and \( P \) such that

\[ \lim\limits_{n \to \infty} s_n = L \mbox{ and } \lim\limits_{n \to \infty} t_n = P . \]

Then the following rules hold:

**Sum Rule:**\[ \lim\limits_{n \to \infty} (s_n + t_n ) = \lim\limits_{n \to \infty} s_n + \lim\limits_{n \to \infty} t_n = L + P . \]

**Difference Rule:**\[ \lim\limits_{n \to \infty} (s_n - t_n ) = \lim\limits_{n \to \infty} s_n - \lim\limits_{n \to \infty} t_n = L - P . \]

**Product Rule:**\[ \lim\limits_{n \to \infty} (s_n \cdot t_n ) = \left( \lim\limits_{n \to \infty} s_n \right) \cdot \left( \lim\limits_{n \to \infty} t_n \right) = L \cdot P . \]

**Constant Multiple Rule:**for any constant \( C \),\[ \lim\limits_{n \to \infty} (C \cdot s_n ) = C\cdot \lim\limits_{n \to \infty} s_n = C \cdot L. \]

**Quotient Rule:**If \( P \not= 0 \) and \( t_n \not= 0 \) for all \( n \in \mathbb{n} \), then\[ \lim\limits_{n \to \infty} \left( \frac{s_n}{t_n} \right) = \frac{\lim\limits_{n \to \infty} s_n }{ \lim\limits_{n \to \infty} t_n }= \frac{L}{P} . \]

Use the Properties of Limits for a Sequence.

More about Limit of a Sequence

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