Sequences and Their Properties: Learn It 3

Giselle Miller

Sequences and Their Properties: Learn It 3

Evaluating Sequence Limits

Now that we understand the basic theory of sequence limits, let’s put our tools to work. We’ll start by analyzing geometric sequences and then move on to more complex examples using algebraic limit laws.

Geometric Sequences and Their Limits

We can use our theorem about function limits to evaluate [latex]\underset{n\to \infty }{\text{lim}}r^n[/latex] for different values of [latex]r[/latex].

Consider the sequence [latex]{(\frac{1}{2})^n}[/latex] and the related exponential function [latex]f(x) = (\frac{1}{2})^x[/latex]. Since [latex]\underset{x\to \infty }{\text{lim}}(\frac{1}{2})^x = 0[/latex], we conclude that the sequence [latex]{(\frac{1}{2})^n}[/latex] converges to 0.

This pattern extends to all geometric sequences:

limits of geometric sequences [latex]{r^n}[/latex]

If [latex]0 < r < 1[/latex]: [latex]r^n \to 0[/latex]
If [latex]r = 1[/latex]: [latex]r^n \to 1[/latex]
If [latex]r > 1[/latex]: [latex]r^n \to \infty[/latex] (diverges)

Working with More Complex Sequences

Now let’s tackle sequences with more complicated terms. Consider the sequence [latex]{(\frac{2}{3})^n + (\frac{1}{4})^n}[/latex].

Rather than analyzing this directly, we can break it down using what we know about simpler sequences. Since both [latex]{(\frac{2}{3})^n}[/latex] and [latex]{(\frac{1}{4})^n}[/latex] converge to 0, their sum should converge to [latex]0 + 0 = 0[/latex].

This intuitive approach works because of algebraic limit laws for sequences, which parallel the limit laws you learned for functions.

algebraic limit laws for sequences

Given sequences [latex]\{{a}_{n}\}[/latex] and [latex]\{{b}_{n}\}[/latex] and any real number [latex]c[/latex], if there exist constants [latex]A[/latex] and [latex]B[/latex] such that [latex]\underset{n\to \infty }{\text{lim}}{a}_{n}=A[/latex] and [latex]\underset{n\to \infty }{\text{lim}}{b}_{n}=B[/latex], then:

Constant: [latex]\underset{n\to \infty }{\text{lim}}c=c[/latex]
Scalar multiplication: [latex]\underset{n\to \infty }{\text{lim}}c{a}_{n}=c\underset{n\to \infty }{\text{lim}}{a}_{n}=cA[/latex]
Addition/Subtraction: [latex]\underset{n\to \infty }{\text{lim}}({a}_{n}\pm {b}_{n})=\underset{n\to \infty }{\text{lim}}{a}_{n}\pm \underset{n\to \infty }{\text{lim}}{b}_{n}=A\pm B[/latex]
Multiplication: [latex]\underset{\text{n}\to \infty }{\text{lim}}({a}_{n}\cdot {b}_{n})=(\underset{n\to \infty }{\text{lim}}{a}_{n})\cdot (\underset{n\to \infty }{\text{lim}}{b}_{n})=A\cdot B[/latex]
Division: [latex]\underset{n\to \infty }{\text{lim}}(\frac{{a}_{n}}{{b}_{n}})=\frac{\underset{n\to \infty }{\text{lim}}{a}_{n}}{\underset{n\to \infty }{\text{lim}}{b}_{n}}=\frac{A}{B}[/latex], provided [latex]B\ne 0[/latex] and each [latex]{b}_{n}\ne 0[/latex].

Proof

We prove part iii.

Let [latex]>0[/latex]. Since [latex]\underset{n\to \infty }{\text{lim}}{a}_{n}=A[/latex], there exists a constant positive integer [latex]{N}_{1}[/latex] such that for all [latex]n\ge {N}_{1}[/latex]. Since [latex]\underset{n\to \infty }{\text{lim}}{b}_{n}=B[/latex], there exists a constant [latex]{N}_{2}[/latex] such that [latex]|{b}_{n}-B|<\frac{\epsilon}{2}[/latex] for all [latex]n\ge {N}_{2}[/latex]. Let [latex]N[/latex] be the largest of [latex]{N}_{1}[/latex] and [latex]{N}_{2}[/latex]. Therefore, for all [latex]n\ge N[/latex],

[latex]|\left({a}_{n}+{b}_{n}\right)\text{-}\left(A+B\right)|\le |{a}_{n}-A|+|{b}_{n}-B|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon[/latex].

[latex]_\blacksquare[/latex]

These laws let you break complex sequences into simpler pieces. Look for:

Terms you can factor out (use scalar multiplication)
Sums or differences of familiar sequences
Products of sequences you already understand

Just like with function limits, these tools make seemingly difficult problems much more manageable.

The algebraic limit laws give us powerful tools for evaluating many sequence limits. Let’s start with a fundamental result that we’ll use repeatedly.

Since we know [latex]\underset{n\to \infty }{\text{lim}}\frac{1}{n} = 0[/latex], we can extend this to any positive power:

Key result: For any positive integer [latex]k[/latex]: [latex]\underset{n\to \infty }{\text{lim}}\frac{1}{n^k} = 0[/latex]

This simple fact, combined with our limit laws, allows us to handle many complex sequences. But first, let’s recall an important technique from precalculus that will help us with rational expressions.

End Behavior of Rational Functions

For [latex]\underset{x\to \infty }{\text{lim}}\frac{P(x)}{Q(x)}[/latex] where [latex]P(x)[/latex] and [latex]Q(x)[/latex] are polynomials:

Case 1: If degree of [latex]P(x) <[/latex] degree of [latex]Q(x)[/latex]: [latex]\underset{x\to \infty }{\text{lim}} \frac{P(x)}{Q(x)} = 0[/latex]

Case 2: If degree of [latex]P(x) >[/latex] degree of [latex]Q(x)[/latex]: [latex]\underset{x\to \infty }{\text{lim}} \frac{P(x)}{Q(x)} = \infty[/latex] (or [latex]-\infty[/latex])

Case 3: If degree of [latex]P(x) =[/latex] degree of [latex]Q(x)[/latex]: [latex]\underset{x\to \infty }{\text{lim}}\frac{P(x)}{Q(x)} = \frac{\text{leading coefficient of }P(x)}{\text{leading coefficient of }Q(x)}[/latex]

For each of the following sequences, determine whether or not the sequence converges. If it converges, find its limit.

[latex]\left\{5-\frac{3}{{n}^{2}}\right\}[/latex]
[latex]\left\{\frac{3{n}^{4}-7{n}^{2}+5}{6 - 4{n}^{4}}\right\}[/latex]
[latex]\left\{\frac{{2}^{n}}{{n}^{2}}\right\}[/latex]
[latex]\left\{{\left(1+\frac{4}{n}\right)}^{n}\right\}[/latex]

We know that [latex]\frac{1}{n}\to 0[/latex]. Using this fact, we conclude that

[latex]\underset{n\to \infty }{\text{lim}}\frac{1}{{n}^{2}}=\underset{n\to \infty }{\text{lim}}\left(\frac{1}{n}\right).\underset{n\to \infty }{\text{lim}}\left(\frac{1}{n}\right)=0[/latex].

Therefore,

[latex]\underset{n\to \infty }{\text{lim}}\left(5-\frac{3}{{n}^{2}}\right)=\underset{n\to \infty }{\text{lim}}5 - 3\underset{n\to \infty }{\text{lim}}\frac{1}{{n}^{2}}=5 - 3.0=5[/latex].

The sequence converges and its limit is [latex]5[/latex].
By factoring [latex]{n}^{4}[/latex] out of the numerator and denominator and using the limit laws above, we have

[latex]\begin{array}{cc}\hfill \underset{n\to \infty }{\text{lim}}\frac{3{n}^{4}-7{n}^{2}+5}{6 - 4{n}^{4}}& =\underset{n\to \infty }{\text{lim}}\frac{3-\frac{7}{{n}^{2}}+\frac{5}{{n}^{4}}}{\frac{6}{{n}^{4}}-4}\hfill \\ & =\frac{\underset{n\to \infty }{\text{lim}}\left(3-\frac{7}{{n}^{2}}+\frac{5}{{n}^{4}}\right)}{\underset{n\to \infty }{\text{lim}}\left(\frac{6}{{n}^{4}}-4\right)}\hfill \\ & =\frac{\left(\underset{n\to \infty }{\text{lim}}\left(3\right)\text{-}\underset{n\to \infty }{\text{lim}}\frac{7}{{n}^{2}}+\underset{n\to \infty }{\text{lim}}\frac{5}{{n}^{4}}\right)}{\left(\underset{n\to \infty }{\text{lim}}\frac{6}{{n}^{4}}-\underset{n\to \infty }{\text{lim}}\left(4\right)\right)}\hfill \\ & =\frac{\left(\underset{n\to \infty }{\text{lim}}\left(3\right)\text{-}7\cdot \underset{n\to \infty }{\text{lim}}\frac{1}{{n}^{2}}+5\cdot \underset{n\to \infty }{\text{lim}}\frac{1}{{n}^{4}}\right)}{\left(6\cdot \underset{n\to \infty }{\text{lim}}\frac{1}{{n}^{4}}-\underset{n\to \infty }{\text{lim}}\left(4\right)\right)}\hfill \\ & =\frac{3 - 7\cdot 0+5\cdot 0}{6\cdot 0 - 4}=-\frac{3}{4}.\hfill \end{array}[/latex]

The sequence converges and its limit is [latex]\frac{-3}{4}[/latex].
Consider the related function [latex]f\left(x\right)=\frac{{2}^{x}}{{x}^{2}}[/latex] defined on all real numbers [latex]x>0[/latex]. Since [latex]{2}^{x}\to \infty[/latex] and [latex]{x}^{2}\to \infty[/latex] as [latex]x\to \infty[/latex], apply L’Hôpital’s rule and write

[latex]\begin{array}{ccccc}\underset{x\to \infty }{\text{lim}}\frac{{2}^{x}}{{x}^{2}}\hfill & =\underset{x\to \infty }{\text{lim}}\frac{{2}^{x}\text{ln}2}{2x}\hfill & & & \text{Take the derivatives of the numerator and denominator.}\hfill \\ & =\underset{x\to \infty }{\text{lim}}\frac{{2}^{x}{\left(\text{ln}2\right)}^{2}}{2}\hfill & & & \text{Take the derivatives again.}\hfill \\ & =\infty .\hfill & & & \end{array}[/latex]

We conclude that the sequence diverges.
Consider the function [latex]f\left(x\right)={\left(1+\frac{4}{x}\right)}^{x}[/latex] defined on all real numbers [latex]x>0[/latex]. This function has the indeterminate form [latex]{1}^{\infty }[/latex] as [latex]x\to \infty[/latex]. Let

[latex]y=\underset{x\to \infty }{\text{lim}}{\left(1+\frac{4}{x}\right)}^{x}[/latex].

Now taking the natural logarithm of both sides of the equation, we obtain

[latex]\text{ln}\left(y\right)=\text{ln}\left[\underset{x\to \infty }{\text{lim}}{\left(1+\frac{4}{x}\right)}^{x}\right][/latex].

Since the function [latex]f\left(x\right)=\text{ln}x[/latex] is continuous on its domain, we can interchange the limit and the natural logarithm. Therefore,

[latex]\text{ln}\left(y\right)=\underset{x\to \infty }{\text{lim}}\left[\text{ln}{\left(1+\frac{4}{x}\right)}^{x}\right][/latex].

Using properties of logarithms, we write

[latex]\underset{x\to \infty }{\text{lim}}\left[\text{ln}{\left(1+\frac{4}{x}\right)}^{x}\right]=\underset{x\to \infty }{\text{lim}}x\text{ln}\left(1+\frac{4}{x}\right)[/latex].

Since the right-hand side of this equation has the indeterminate form [latex]\infty \cdot 0[/latex], rewrite it as a fraction to apply L’Hôpital’s rule. Write

[latex]\underset{x\to \infty }{\text{lim}}x\text{ln}\left(1+\frac{4}{x}\right)=\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\left(1+\frac{4}{x}\right)}{\frac{1}{x}}[/latex].

Since the right-hand side is now in the indeterminate form [latex]\frac{0}{0}[/latex], we are able to apply L’Hôpital’s rule. We conclude that

[latex]\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\left(1+\frac{4}{x}\right)}{\frac{1}{x}}=\underset{x\to \infty }{\text{lim}}\frac{4}{1+\frac{4}{x}}=4[/latex].

Therefore, [latex]\text{ln}\left(y\right)=4[/latex] and [latex]y={e}^{4}[/latex]. Therefore, since [latex]\underset{x\to \infty }{\text{lim}}{\left(1+\frac{4}{x}\right)}^{x}={e}^{4}[/latex], we can conclude that the sequence [latex]\left\{{\left(1+\frac{4}{n}\right)}^{n}\right\}[/latex] converges to [latex]{e}^{4}[/latex].