2.1 Sequences and limits (Page 2/2)

Analysis of functions of a single Page 2 / 2

It is clear from this definition that we can't check whether a sequence converges or not unless we know thelimit value $L .$ The whole thrust of this definition has to do with estimating the quantity $| a_{n} - L | .$ We will see later that there are ways to tell in advance that a sequence converges without knowing the value of the limit.

Let $a_{n} = 1 / n,$ and let us show that $lim a_{n} = 0 .$ Given an $ϵ > 0,$ let us choose a $N$ such that $1 / N < ϵ .$ (How do we know we can find such a $N ?$ ) Now, if $n \geq N,$ then we have

| a_{n} - 0 | = | \frac{1}{n} | = \frac{1}{n} \leq \frac{1}{N} < ϵ,

which is exactly what we needed to show to conclude that $0 = lim a_{n} .$

Let $a_{n} = (2 n + 1) / (1 - 3 n),$ and let $L = - 2 / 3 .$ Let us show that $L = lim a_{n} .$ Indeed, if $ϵ > 0$ is given, we must find a $N,$ such that if $n \geq N$ then $| a_{n} + (2 / 3) | < ϵ .$ Let us examine the quantity $| a_{n} + 2 / 3 | .$ Maybe we can make some estimates on it, in such a way that it becomes clear how to find the natural number $N .$

\begin{matrix} | a_{n} + (2 / 3) | & = & | \frac{2 n + 1}{1 - 3 n} + \frac{2}{3} | \\ = & | \frac{6 n + 3 + 2 - 6 n}{3 - 9 n} | \\ = & | \frac{5}{3 - 9 n} | \\ = & \frac{5}{9 n - 3} \\ = & \frac{5}{6 n + 3 n - 3} \\ \leq & \frac{5}{6 n} \\ < & \frac{1}{n}, \end{matrix}

for all $n \geq 1 .$ Therefore, if $N$ is an integer for which $N > 1 / ϵ,$ then

| a_{n} + 2 / 3 | < 1 / n \leq 1 / N < ϵ,

whenever $n \geq N,$ as desired. (How do we know that there exists a $N$ which is larger than the number $1 / ϵ ?$ )

Let $a_{n} = 1 / \sqrt{n},$ and let us show that $lim a_{n} = 0 .$ Given an $ϵ > 0,$ we must find an integer $N$ that satisfies the requirements of the definition. It's a little trickier this time to choose this $N .$ Consider the positive number $ϵ^{2} .$ We know, from Exercise 1.16, that there exists a natural number $N$ such that $1 / N < ϵ^{2} .$ Now, if $n \geq N,$ then

| a_{n} - 0 | = \frac{1}{\sqrt{n}} \leq \frac{1}{\sqrt{N}} = \sqrt{\frac{1}{N}} < \sqrt{ϵ^{2}} = ϵ,

which shows that $0 = lim 1 / \sqrt{n} .$

REMARK A good way to attack a limit problem is to immediately examine the quantity $| a_{n} - L |,$ which is what we did in [link] above. This is the quantity we eventually wish to show is less than $ϵ$ when $n \geq N,$ and determining which $N$ to use is always the hard part. Ordinarily, some algebraic manipulations can be performed on the expression $| a_{n} - L |$ that can help us figure out exactly how to choose $N .$ Just know that this process takes some getting used to, so practice!

Using the basic definition, prove that $lim 3 / (2 n + 7) = 0 .$
Using the basic definition, prove that $lim 1 / n^{2} = 0 .$
Using the basic definition, prove that $lim (n^{2} + 1) / (n^{2} + 100 n) = 1 .$ HINT: Use the idea from the remark above; i.e., examine the quantity $| a_{n} - L | .$
Again, using the basic definition, prove that
$lim \frac{n + n^{2} i}{n - n^{2} i} = - 1 .$
Remember the definition of the absolute value of a complex number.
Using the basic definition, prove that
$lim \frac{n^{3} + n^{2} i}{1 - n^{3} i} = i .$
Let $a_{n} = {(- 1)}^{n} .$ Prove that 1 is not the limit of the sequence ${a_{n}} .$ HINT: Suppose the sequence ${a_{n}}$ does converge to 1. Use $ϵ = 1,$ let $N$ be the corresponding integer that exists in the definition, satisfying $| a_{n} - 1 | < 1$ for all $n \geq N,$ and then examine the quantity $| a_{n} - 1 |$ for various $n$ 's to get a contradiction.

Let ${a_{n}}$ be a sequence of (real or complex) numbers, and let $L$ be a number. Prove that $L = lim a_{n}$ if and only if for every positive integer $k$ there exists an integer $N,$ such that if $n \geq N$ then $| a_{n} - L | < 1 / k .$
Let ${c_{n}}$ be a sequence of complex numbers, and suppose that $c_{n} \mapsto L .$ If $c_{n} = a_{n} + b_{n} i$ and $L = a + b i,$ show that $a = lim a_{n}$ and $b = lim b_{n} .$ Conversely, if $a = lim a_{n}$ and $b = lim b_{n},$ show that $a + b i = lim (a_{n} + b_{n} i) .$ That is, a sequence ${c_{n} = a_{n} + b_{n} i}$ of complex numbers converges if and only if the sequence ${a_{n}}$ of the real parts converges and the sequence ${b_{n}}$ of the imaginary parts converges.HINT: You need to show that, given some hypotheses, certain quantities are less than $ϵ .$ Part (c) of [link] should be of help.

Prove that a constant sequence ( $a_{n} \equiv c$ ) converges to $c .$
Prove that the sequence ${\frac{2 n^{2} + 1}{1 - 3 n}}$ diverges to $- \infty .$
Prove that the sequence ${{(- 1)}^{n}}$ does not converge to any number $L .$ HINT: Argue by contradiction. Suppose it does converge to a number $L .$ Use $ϵ = 1 / 2,$ let $N$ be the corresponding integer that exists in the definition, and then examine $| a_{n} - a_{n + 1} |$ for $n \geq N .$ Use the following useful add and subtract trick:
$| a_{n} - a_{n + 1} | = | a_{n} - L + L - a_{n + 1} | \leq | a_{n} - L | + | L - a_{n + 1} | .$

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Source: OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1

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