5.1 Sequences (Page 6/25)

Calculus volume 2 Page 6 / 25

lim_{n \to \infty} \sqrt{5 - \frac{3}{n^{2}}} = \sqrt{lim_{n \to \infty} (5 - \frac{3}{n^{2}})} = \sqrt{5} .

Continuous functions defined on convergent sequences

Consider a sequence ${a_{n}}$ and suppose there exists a real number $L$ such that the sequence ${a_{n}}$ converges to $L .$ Suppose $f$ is a continuous function at $L .$ Then there exists an integer $N$ such that $f$ is defined at all values $a_{n}$ for $n \geq N,$ and the sequence ${f (a_{n})}$ converges to $f (L)$ ( [link] ).

Proof

Let $ϵ > 0 .$ Since $f$ is continuous at $L,$ there exists $δ > 0$ such that $| f (x) - f (L) | < ε$ if $| x - L | < δ .$ Since the sequence ${a_{n}}$ converges to $L,$ there exists $N$ such that $| a_{n} - L | < δ$ for all $n \geq N .$ Therefore, for all $n \geq N,$ $| a_{n} - L | < δ,$ which implies $| f (a_{n}) − f (L) | < ε .$ We conclude that the sequence ${f (a_{n})}$ converges to $f (L) .$

□

A graph in quadrant 1 with points (a_1, f(a_1)), (a_3, f(a_3)), (L, f(L)), (a_4, f(a_4)), and (a_2, f(a_2)) connected by smooth curves. — Because $f$ is a continuous function as the inputs $a_{1}, a_{2}, a_{3},…$ approach $L,$ the outputs $f (a_{1}), f (a_{2}), f (a_{3}),…$ approach $f (L) .$

Limits involving continuous functions defined on convergent sequences

Determine whether the sequence ${cos (3 / n^{2})}$ converges. If it converges, find its limit.

Since the sequence ${3 / n^{2}}$ converges to $0$ and $cos x$ is continuous at $x = 0,$ we can conclude that the sequence ${cos (3 / n^{2})}$ converges and

lim_{n \to \infty} cos (\frac{3}{n^{2}}) = cos (0) = 1 .

Got questions? Get instant answers now!

Determine if the sequence ${\sqrt{\frac{2 n + 1}{3 n + 5}}}$ converges. If it converges, find its limit.

The sequence converges, and its limit is $\sqrt{2 / 3} .$

Got questions? Get instant answers now!

Another theorem involving limits of sequences is an extension of the Squeeze Theorem for limits discussed in Introduction to Limits .

Squeeze theorem for sequences

Consider sequences ${a_{n}},$ ${b_{n}},$ and ${c_{n}} .$ Suppose there exists an integer $N$ such that

a_{n} \leq b_{n} \leq c_{n} for all n \geq N .

If there exists a real number $L$ such that

lim_{n \to \infty} a_{n} = L = lim_{n \to \infty} c_{n},

then ${b_{n}}$ converges and $lim_{n \to \infty} b_{n} = L$ ( [link] ).

Proof

Let $ε > 0 .$ Since the sequence ${a_{n}}$ converges to $L,$ there exists an integer $N_{1}$ such that $| a_{n} - L | < ε$ for all $n \geq N_{1} .$ Similarly, since ${c_{n}}$ converges to $L,$ there exists an integer $N_{2}$ such that $| c_{n} - L | < ε$ for all $n \geq N_{2} .$ By assumption, there exists an integer $N$ such that $a_{n} \leq b_{n} \leq c_{n}$ for all $n \geq N .$ Let $M$ be the largest of $N_{1}, N_{2},$ and $N .$ We must show that $| b_{n} - L | < ε$ for all $n \geq M .$ For all $n \geq M,$

− ε < − | a_{n} - L | \leq a_{n} - L \leq b_{n} - L \leq c_{n} - L \leq | c_{n} - L | < ε .

Therefore, $− ε < b_{n} - L < ε,$ and we conclude that $| b_{n} - L | < ε$ for all $n \geq M,$ and we conclude that the sequence ${b_{n}}$ converges to $L .$

□

A graph in quadrant 1 with the line y = L and the x axis labeled as the n axis. Points are plotted above and below the line, converging to L as n goes to infinity. Points a_n, b_n, and c_n are plotted at the same n-value. A_n and b_n are above y = L, and c_n is below it. — Each term $b_{n}$ satisfies $a_{n} \leq b_{n} \leq c_{n}$ and the sequences ${a_{n}}$ and ${c_{n}}$ converge to the same limit, so the sequence ${b_{n}}$ must converge to the same limit as well.

Using the squeeze theorem

Use the Squeeze Theorem to find the limit of each of the following sequences.

${\frac{cos n}{n^{2}}}$
${{(- \frac{1}{2})}^{n}}$

Since $−1 \leq cos n \leq 1$ for all integers $n,$ we have

$- \frac{1}{n^{2}} \leq \frac{cos n}{n^{2}} \leq \frac{1}{n^{2}} .$

Since $−1 / n^{2} \to 0$ and $1 / n^{2} \to 0,$ we conclude that $cos n / n^{2} \to 0$ as well.
Since

$- \frac{1}{2^{n}} \leq {(- \frac{1}{2})}^{n} \leq \frac{1}{2^{n}}$

for all positive integers $n,$ $−1 / 2^{n} \to 0$ and $1 / 2^{n} \to 0,$ we can conclude that ${(−1 / 2)}^{n} \to 0 .$

Got questions? Get instant answers now!

Find $lim_{n \to \infty} \frac{2 n - sin n}{n} .$

$2$

Got questions? Get instant answers now!

Using the idea from [link] b. we conclude that $r^{n} \to 0$ for any real number $r$ such that $−1 < r < 0 .$ If $r < − 1,$ the sequence ${r^{n}}$ diverges because the terms oscillate and become arbitrarily large in magnitude. If $r = −1,$ the sequence ${r^{n}} = {{(−1)}^{n}}$ diverges, as discussed earlier. Here is a summary of the properties for geometric sequences.

r^{n} \to 0 if | r | < 1

r^{n} \to 1 if r = 1

r^{n} \to \infty if r > 1

{r^{n}} diverges if r \leq − 1

Bounded sequences

We now turn our attention to one of the most important theorems involving sequences: the Monotone Convergence Theorem. Before stating the theorem, we need to introduce some terminology and motivation. We begin by defining what it means for a sequence to be bounded.

<< Chapter < Page Page > Chapter >>

Practice Key Terms 15

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 2' conversation and receive update notifications?

Ask

	3 Neuroscience Exam 2004 4 By David Corey Start Exam
	Summer kitchens By Heather McAvoy Start Quiz
	Music 113 By Eric Crawford Start Quiz
	26 AP Key Terms 26 Fluid, Electrolyte, Acid-Base By OpenStax Start Key Terms
	15 AP 15 Autonomic Nervous System Essay By OpenStax Start Flashcards
	Job Search Skills MCQ By Mary Matera Start Quiz
	1 Psychology Concept Test By John Gabrieli Start Test
	Subject-verb Agreement By Dindin Secreto Start Quiz
	8 Arts Society: Theater 8 By Jonathan Long Start Quiz
©flickr: Luis	Atoms By Carly Allen Start Quiz