5.1 Sequences (Page 8/25)

Calculus volume 2 Page 8 / 25

A graph in quadrant 1 with the x and y axes labeled n and a_n, respectively. A dotted horizontal is drawn from the a_n axis into quadrant 1. Many points are plotted under the dotted line, increasing in a_n value and converging to the dotted line. — Since the sequence ${a_{n}}$ is increasing and bounded above, it must converge.

In the following example, we show how the Monotone Convergence Theorem can be used to prove convergence of a sequence.

Using the monotone convergence theorem

For each of the following sequences, use the Monotone Convergence Theorem to show the sequence converges and find its limit.

${\frac{4^{n}}{n!}}$
${a_{n}}$ defined recursively such that

$a_{1} = 2 and a_{n + 1} = \frac{a_{n}}{2} + \frac{1}{2 a_{n}} for all n \geq 2 .$

Writing out the first few terms, we see that

${\frac{4^{n}}{n!}} = {4, 8, \frac{32}{3}, \frac{32}{3}, \frac{128}{15},…} .$

At first, the terms increase. However, after the third term, the terms decrease. In fact, the terms decrease for all $n \geq 3 .$ We can show this as follows.

$a_{n + 1} = \frac{4^{n + 1}}{(n + 1)!} = \frac{4}{n + 1} \cdot \frac{4^{n}}{n!} = \frac{4}{n + 1} \cdot a_{n} \leq a_{n} i f n \geq 3 .$

Therefore, the sequence is decreasing for all $n \geq 3 .$ Further, the sequence is bounded below by $0$ because $4^{n} / n! \geq 0$ for all positive integers $n .$ Therefore, by the Monotone Convergence Theorem, the sequence converges.
To find the limit, we use the fact that the sequence converges and let $L = lim_{n \to \infty} a_{n} .$ Now note this important observation. Consider $lim_{n \to \infty} a_{n + 1} .$ Since

${a_{n + 1}} = {a_{2,} a_{3}, a_{4},…},$
the only difference between the sequences ${a_{n + 1}}$ and ${a_{n}}$ is that ${a_{n + 1}}$ omits the first term. Since a finite number of terms does not affect the convergence of a sequence,

$lim_{n \to \infty} a_{n + 1} = lim_{n \to \infty} a_{n} = L .$

Combining this fact with the equation

$a_{n + 1} = \frac{4}{n + 1} a_{n}$

and taking the limit of both sides of the equation

$lim_{n \to \infty} a_{n + 1} = lim_{n \to \infty} \frac{4}{n + 1} a_{n},$

we can conclude that

$L = 0 \cdot L = 0 .$
Writing out the first several terms,

${2, \frac{5}{4}, \frac{41}{40}, \frac{3281}{3280},…} .$

we can conjecture that the sequence is decreasing and bounded below by $1 .$ To show that the sequence is bounded below by $1,$ we can show that

$\frac{a_{n}}{2} + \frac{1}{2 a_{n}} \geq 1 .$

To show this, first rewrite

$\frac{a_{n}}{2} + \frac{1}{2 a_{n}} = \frac{a_{n}^{2} + 1}{2 a_{n}} .$

Since $a_{1} > 0$ and $a_{2}$ is defined as a sum of positive terms, $a_{2} > 0 .$ Similarly, all terms $a_{n} > 0 .$ Therefore,

$\frac{a_{n}^{2} + 1}{2 a_{n}} \geq 1$

if and only if

$a_{n}^{2} + 1 \geq 2 a_{n} .$

Rewriting the inequality $a_{n}^{2} + 1 \geq 2 a_{n}$ as $a_{n}^{2} - 2 a_{n} + 1 \geq 0,$ and using the fact that

$a_{n}^{2} - 2 a_{n} + 1 = {(a_{n} - 1)}^{2} \geq 0$

because the square of any real number is nonnegative, we can conclude that

$\frac{a_{n}}{2} + \frac{1}{2 a_{n}} \geq 1 .$

To show that the sequence is decreasing, we must show that $a_{n + 1} \leq a_{n}$ for all $n \geq 1 .$ Since $1 \leq a_{n}^{2},$ it follows that

$a_{n}^{2} + 1 \leq 2 a_{n}^{2} .$

Dividing both sides by $2 a_{n},$ we obtain

$\frac{a_{n}}{2} + \frac{1}{2 a_{n}} \leq a_{n} .$

Using the definition of $a_{n + 1},$ we conclude that

$a_{n + 1} = \frac{a_{n}}{2} + \frac{1}{2 a_{n}} \leq a_{n} .$

Since ${a_{n}}$ is bounded below and decreasing, by the Monotone Convergence Theorem, it converges.
To find the limit, let $L = lim_{n \to \infty} a_{n} .$ Then using the recurrence relation and the fact that $lim_{n \to \infty} a_{n} = lim_{n \to \infty} a_{n + 1},$ we have

$lim_{n \to \infty} a_{n + 1} = lim_{n \to \infty} (\frac{a_{n}}{2} + \frac{1}{2 a_{n}}),$

and therefore

$L = \frac{L}{2} + \frac{1}{2 L} .$

Multiplying both sides of this equation by $2 L,$ we arrive at the equation

$2 L^{2} = L^{2} + 1 .$

Solving this equation for $L,$ we conclude that $L^{2} = 1,$ which implies $L = ± 1 .$ Since all the terms are positive, the limit $L = 1 .$

Got questions? Get instant answers now!

Consider the sequence ${a_{n}}$ defined recursively such that $a_{1} = 1,$ $a_{n} = a_{n - 1} / 2 .$ Use the Monotone Convergence Theorem to show that this sequence converges and find its limit.

$0 .$

Got questions? Get instant answers now!

Fibonacci numbers

The Fibonacci numbers are defined recursively by the sequence ${F_{n}}$ where $F_{0} = 0,$ $F_{1} = 1$ and for $n \geq 2,$

F_{n} = F_{n - 1} + F_{n - 2} .

Here we look at properties of the Fibonacci numbers.

Write out the first twenty Fibonacci numbers.
Find a closed formula for the Fibonacci sequence by using the following steps.
1. Consider the recursively defined sequence ${x_{n}}$ where $x_{o} = c$ and $x_{n + 1} = a x_{n} .$ Show that this sequence can be described by the closed formula $x_{n} = c a^{n}$ for all $n \geq 0 .$
2. Using the result from part a. as motivation, look for a solution of the equation
  
  $F_{n} = F_{n - 1} + F_{n - 2}$
  
  of the form $F_{n} = c λ^{n} .$ Determine what two values for $λ$ will allow $F_{n}$ to satisfy this equation.
3. Consider the two solutions from part b.: $λ_{1}$ and $λ_{2} .$ Let $F_{n} = c_{1} λ_{1}^{n} + c_{2} λ_{2}^{n} .$ Use the initial conditions $F_{0}$ and $F_{1}$ to determine the values for the constants $c_{1}$ and $c_{2}$ and write the closed formula $F_{n} .$
Use the answer in 2 c. to show that

$lim_{n \to \infty} \frac{F_{n + 1}}{F_{n}} = \frac{1 + \sqrt{5}}{2} .$

The number $ϕ = (1 + \sqrt{5}) / 2$ is known as the golden ratio ( [link] and [link] ).

The seeds in a sunflower exhibit spiral patterns curving to the left and to the right. The number of spirals in each direction is always a Fibonacci number—always. (credit: modification of work by Esdras Calderan, Wikimedia Commons)

The proportion of the golden ratio appears in many famous examples of art and architecture. The ancient Greek temple known as the Parthenon was designed with these proportions, and the ratio appears again in many of the smaller details. (credit: modification of work by TravelingOtter, Flickr)

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

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