2.4 Properties of convergent sequences

Analysis of functions of a single Page 1 / 1

A discussion of the properties of convergent sequences, such as when they form a bounded set, the squeeze theorem, and the properties of convergence when sequences are combined algebraically.

Often, our goal is to show that a given sequence is convergent. However, as we study convergent sequences, we would liketo establish various properties that they have in common. The first theorem of this section is just such a result.

Suppose ${a_{n}}$ is a convergent sequence of real or complex numbers. Then the sequence ${a_{n}}$ forms a bounded set.

Write $L = lim a_{n} .$ Let $ϵ$ be the positive number $1 .$ Then, there exists a natural number $N$ such that $| a_{n} - L | < 1$ for all $n \geq N .$ By the backward triangle inequality, this implies that $| | a_{n} | - | L | | < 1$ for all $n \geq N,$ which implies that $| a_{n} | \leq | L | + 1$ for all $n \geq N .$ This shows that at least the tail of the sequence is bounded by the constant $| L | + 1 .$

Next, let $K$ be a number larger than the finitely many numbers $| a_{1} |, ..., | a_{N - 1} | .$ Then, for any $n,$ $| a_{n} |$ is either less than $K$ or $| L | + 1 .$ Let $M$ be the larger of the two numbers $K$ and $| L | + 1 .$ Then $| a_{n} | < M$ for all $n .$ Hence, the sequence ${a_{n}}$ is bounded.

Note that the preceding theorem is a partial converse to [link] ; i.e., a convergent sequence is necessarily bounded. Of course, not every convergent sequence must beeither nondecreasing or nonincreasing, so that a full converse to [link] is not true. For instance, take $z = - 1 / 2$ in part (1) of [link] . It converges to 0 all right, but it is neither nondecreasing nor nonincreasing.

Suppose ${a_{n}}$ is a sequence of real numbers that converges to a number $a,$ and assume that $a_{n} \geq c$ for all $n .$ Prove that $a \geq c .$ HINT: Suppose not, and let $ϵ$ be the positive number $c - a .$ Let $N$ be a natural number corresponding to this choice of $ϵ,$ and derive a contradiction.
If ${a_{n}}$ is a sequence of real numbers for which $lim a_{n} = a,$ and if $a \neq 0,$ then prove that $a_{n} \neq 0$ for all large enough $n .$ Show in fact that there exists an $N$ such that $| a_{n} | > | a | / 2$ for all $n \geq N .$ HINT: Make use of the positive number $ϵ = | a | / 2 .$

If ${a_{n}}$ is a sequence of positive real numbers for which $lim a_{n} = a > 0,$ prove that $lim \sqrt{a_{n}} = \sqrt{a} .$ HINT: Multiply the expression $\sqrt{a_{n}} - \sqrt{a}$ above and below by $\sqrt{a_{n}} + \sqrt{a} .$
If ${a_{n}}$ is a sequence of complex numbers, and $lim a_{n} = a,$ prove that $lim | a_{n} | = | a | .$ HINT: Use the backward triangle inequality.

Suppose ${a_{n}}$ is a sequence of real numbers and that $L = lim a_{n} .$ Let $M_{1}$ and $M_{2}$ be real numbers such that $M_{1} \leq a_{n} \leq M_{2}$ for all $n .$ Prove that $M_{1} \leq L \leq M_{2} .$

HINT: Suppose, for instance, that $L > M_{2} .$ Make use of the positive number $L - M_{2}$ to derive a contradiction.

We are often able to show that a sequence converges by comparing it to another sequence that we already know converges.The following exercise demonstrates some of these techniques.

Let ${a_{n}}$ be a sequence of complex numbers.

Suppose that, for each $n,$ $| a_{n} | < 1 / n .$ Prove that $0 = lim a_{n} .$
Suppose ${b_{n}}$ is a sequence that converges to $0,$ and suppose that, for each $n,$ $| a_{n} | < | b_{n} | .$ Prove that $0 = lim a_{n} .$

The next result is perhaps the most powerful techniquewe have for showing that a given sequence converges to a given number.

Squeeze theorem

Suppose that ${a_{n}}$ is a sequence of real numbers and that ${b_{n}}$ and ${c_{n}}$ are two sequences ofreal numbers for which $b_{n} \leq a_{n} \leq c_{n}$ for all $n .$ Suppose further that $lim b_{n} = lim c_{n} = L .$ Then the sequence ${a_{n}}$ also converges to $L .$

We examine the quantity $| a_{n} - L, |$ employ some add and subtract tricks, and make the following computations:

\begin{matrix} r c l | a_{n} - L | & \leq & | a_{n} - b_{n} + b_{n} - L | \\ \leq & | a_{n} - b_{n} | + | b_{n} - L | \\ = & a_{n} - b_{n} + | b_{n} - L | \\ \leq & c_{n} - b_{n} + | b_{n} - L | \\ = & | c_{n} - b_{n} | + | b_{n} - L | \\ \leq & | c_{n} - L | + | L - b_{n} | + | b_{n} - L | . \end{matrix}

So, we can make $| a_{n} - L | < ϵ$ by making $| c_{n} - L | < ϵ / 3$ and $| b_{n} - L | < ϵ / 3 .$ So, let $N_{1}$ be a positive integer such that $| c_{n} - L | < ϵ / 3$ if $n \geq N_{1},$ and let $N_{2}$ be a positive integer so that $| b_{n} - L | < ϵ / 3$ if $n \geq N_{2} .$ Then set $N = max (N_{1}, N_{2}) .$ Clearly, if $n \geq N,$ then both inequalities $| c_{n} - L | < ϵ / 3$ and $| b_{n} - L | < ϵ / 3,$ and hence $| a_{n} - L | < ϵ .$ This finishes the proof.

The next result establishes what are frequently called the “limit theorems.” Basically, these results show how convergenceinteracts with algebraic operations.

Let ${a_{n}}$ and ${b_{n}}$ be two sequences of complex numbers with $a = lim a_{n}$ and $b = lim b_{n} .$ Then

The sequence ${a_{n} + b_{n}}$ converges, and
$lim (a_{n} + b_{n}) = lim a_{n} + lim b_{n} = a + b .$
The sequence ${a_{n} b_{n}}$ is convergent, and
$lim (a_{n} b_{n}) = lim a_{n} lim b_{n} = a b .$
If all the $b_{n}$ 's as well as $b$ are nonzero, then the sequence ${a_{n} / b_{n}}$ is convergent, and
$lim (\frac{a_{n}}{b_{n}} = \frac{lim a_{n}}{lim b_{n}} = \frac{a}{b} .$

Part (1) is exactly the same as [link] . Let us prove part (2).

By [link] , both sequences ${a_{n}}$ and ${b_{n}}$ are bounded. Therefore, let $M$ be a number such that $| a_{n} | \leq M$ and $| b_{n} | \leq M$ for all $n .$ Now, let $ϵ > 0$ be given. There exists an $N_{1}$ such that $| a_{n} - a | < ϵ / (2 M)$ whenever $n \geq N_{1},$ and there exists an $N_{2}$ such that $| b_{n} - b | < ϵ / (2 M)$ whenever $n \geq N_{2} .$ Let $N$ be the maximum of $N_{1}$ and $N_{2} .$ Here comes the add and subtract trick again.

\begin{matrix} r c l | a_{n} b_{n} - a b | & = & | a_{n} b_{n} - a b_{n} + a b_{n} - a b | \\ \leq & | a_{n} b_{n} - a b_{n} | + | a b_{n} - a b | \\ = & | a_{n} - a | | b_{n} | + | a | | b - b_{n} | \\ \leq & | a_{n} - a | M + M | b_{n} - b | \\ < & ϵ \end{matrix}

if $n \geq N,$ which shows that $lim (a_{n} b_{n}) = a b .$

To prove part (3), let $M$ be as in the previous paragraph, and let $ϵ > 0$ be given. There exists an $N_{1}$ such that $| a_{n} - {a | < (ϵ | b |}^{2}) / (4 M)$ whenever $n \geq N_{1};$ there also exists an $N_{2}$ such that $| b_{n} - {b | < (ϵ | b |}^{2}) / (4 M)$ whenever $n \geq N_{2};$ and there exists an $N_{3}$ such that $| b_{n} | > | b | / 2$ whenever $n \geq N_{3} .$ (See [link] .) Let $N$ be the maximum of the three numbers $N_{1}, N_{2}$ and $N_{3} .$ Then:

\begin{matrix} r c l | \frac{a_{n}}{b_{n}} - \frac{a}{b} | & = & | \frac{a_{n} b - b_{n} a}{b_{n} b} | \\ = & | a_{n} b - b_{n} a | \frac{1}{| b_{n} b |} \\ < & | a_{n} b - b_{n} a | \frac{1}{{| b |}^{2} / 2} \\ \leq & (| a_{n} - a | | b | + | a | | b_{n} - b |) \frac{2}{{| b |}^{2}} \\ < & (M | a_{n} - a | + M | b_{n} - b |) \frac{2}{{| b |}^{2}} \\ < & ϵ \end{matrix}

if $n \geq N .$ This completes the proof.

REMARK The proof of part (3) of the preceding theorem may look mysterious. Where, for instance, does this number ${ϵ | b |}^{2} / 4 M$ come from? The answer is that one begins such a proof by examining the quantity $| a_{n} / b_{n} - a / b |$ to see if by some algebraic manipulation one can discover how to control its size by using the quantities $| a_{n} - a |$ and $| b_{n} - b | .$ The assumption that $a = lim a_{n}$ and $b = lim b_{n}$ mean exactly that the quantities $| a_{n} - a |$ and $| b_{n} - b |$ can be controlled by requiring $n$ to be large enough. The algebraic computation in the proof above shows that

| \frac{a_{n}}{b_{n}} - \frac{a}{b} | \leq (M | a_{n} - a | + M | b_{n} - b |) \frac{2}{{| b |}^{2}},

and one can then see exactly how small to make $| a_{n} - a |$ and $| b_{n} - b |$ so that $| a_{n} / b_{n} - a / b | < ϵ .$ Indeed, this is the way most limit proofs work.

If possible, determine the limits of the following sequences by using [link] , [link] , [link] , and the squeeze theorem [link] .

${n^{1 / n^{2}}} .$
${{(n^{2})}^{1 / n}} .$
${{(1 + n)}^{1 / n}} .$
${{(1 + n^{2})}^{1 / n^{3}}} .$
${{(1 + 1 / n)}^{2 / n}} .$
${{(1 + 1 / n)}^{2 n}} .$
${{(1 + 1 / n)}^{n^{2}}} .$
${{(1 - 1 / n)}^{n}} .$ HINT: Note that
$1 - 1 / n = \frac{n - 1}{n} = \frac{1}{\frac{n}{n - 1}} = \frac{1}{\frac{n - 1 + 1}{n - 1}} = \frac{1}{1 + \frac{1}{n - 1}} .$
${{(1 - 1 / (2 n))}^{3 n}} .$
${{(n!)}^{1 / n}} .$

Questions & Answers

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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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