RA Facts about limits of sequences

Section 2.2 Facts about limits of sequences

Note: 2–2.5 lectures, recursively defined sequences can safely be skipped

In this section we go over some basic results about the limits of sequences. We start by looking at how sequences interact with inequalities.

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Subsection 2.2.1 Limits and inequalities

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A basic lemma about limits and inequalities is the so-called squeeze lemma. It allows us to show convergence of sequences in difficult cases if we find two other simpler convergent sequences that “squeeze” the original sequence.

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Lemma 2.2.1. Squeeze lemma.

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Let

{a_{n}}_{n = 1}^{\infty},

{b_{n}}_{n = 1}^{\infty},

and

{x_{n}}_{n = 1}^{\infty}

be sequences such that

a_{n} \leq x_{n} \leq b_{n} for all n \in N .

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Suppose

{a_{n}}_{n = 1}^{\infty}

and

{b_{n}}_{n = 1}^{\infty}

converge and

lim_{n \to \infty} a_{n} = lim_{n \to \infty} b_{n} .

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Then

{x_{n}}_{n = 1}^{\infty}

converges and

lim_{n \to \infty} x_{n} = lim_{n \to \infty} a_{n} = lim_{n \to \infty} b_{n} .

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Proof.

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Let

x := lim_{n \to \infty} a_{n} = lim_{n \to \infty} b_{n} .

Let

ϵ > 0

be given. Find an

M_{1}

such that for all

n \geq M_{1},

we have that

| a_{n} - x | < ϵ,

and an

M_{2}

such that for all

n \geq M_{2},

we have

| b_{n} - x | < ϵ .

Set

M := max {M_{1}, M_{2}} .

Suppose

n \geq M .

In particular,

x - a_{n} < ϵ,

x - ϵ < a_{n} .

Similarly,

b_{n} < x + ϵ .

Putting everything together, we find

x - ϵ < a_{n} \leq x_{n} \leq b_{n} < x + ϵ .

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In other words,

- ϵ < x_{n} - x < ϵ

| x_{n} - x | < ϵ .

{x_{n}}_{n = 1}^{\infty}

converges to

x .

See Figure 2.3.

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Figure 2.3. Squeeze lemma proof in picture.

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Example 2.2.2.

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One application of the squeeze lemma is to compute limits of sequences using limits that we already know. For example, consider the sequence

{\frac{1}{n \sqrt{n}}}_{n = 1}^{\infty} .

Since

\sqrt{n} \geq 1

for all

n \in N,

we have

0 \leq \frac{1}{n \sqrt{n}} \leq \frac{1}{n}

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for all

n \in N .

We already know

lim_{n \to \infty}^{1} /_{n} = 0 .

Hence, using the constant sequence

{0}_{n = 1}^{\infty}

and the sequence

{^{1} /_{n}}_{n = 1}^{\infty}

in the squeeze lemma, we conclude

lim_{n \to \infty} \frac{1}{n \sqrt{n}} = 0 .

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Limits, when they exist, preserve non-strict inequalities.

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Lemma 2.2.3.

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Let

{x_{n}}_{n = 1}^{\infty}

and

{y_{n}}_{n = 1}^{\infty}

be convergent sequences and

x_{n} \leq y_{n} for all n \in N .

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Then

lim_{n \to \infty} x_{n} \leq lim_{n \to \infty} y_{n} .

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Proof.

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Let

x := lim_{n \to \infty} x_{n}

and

y := lim_{n \to \infty} y_{n} .

Let

ϵ > 0

be given. Find an

M_{1}

such that for all

n \geq M_{1},

we have

| x_{n} - x | <^{ϵ} /_{2} .

Find an

M_{2}

such that for all

n \geq M_{2},

we have

| y_{n} - y | <^{ϵ} /_{2} .

In particular, for some

n \geq max {M_{1}, M_{2}},

we have

x - x_{n} <^{ϵ} /_{2}

and

y_{n} - y <^{ϵ} /_{2} .

We add these inequalities to obtain

y_{n} - x_{n} + x - y < ϵ, or y_{n} - x_{n} < y - x + ϵ .

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Since

x_{n} \leq y_{n},

we have

0 \leq y_{n} - x_{n}

and hence

0 < y - x + ϵ .

In other words,

x - y < ϵ .

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Because

ϵ > 0

was arbitrary, we obtain

x - y \leq 0 .

Therefore,

x \leq y .

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The next corollary follows by using constant sequences in Lemma 2.2.3. The proof is left as an exercise.

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Corollary 2.2.4.

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If ${x_{n}}_{n = 1}^{\infty}$ is a convergent sequence such that $x_{n} \geq 0$ for all $n \in N,$ then

$lim_{n \to \infty} x_{n} \geq 0.$
Let $a, b \in R$ and let ${x_{n}}_{n = 1}^{\infty}$ be a convergent sequence such that

$a \leq x_{n} \leq b for all n \in N .$

Then

$a \leq lim_{n \to \infty} x_{n} \leq b .$

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In Lemma 2.2.3 and Corollary 2.2.4 we cannot simply replace all the non-strict inequalities with strict inequalities. For example, let

x_{n} :=^{- 1} /_{n}

and

y_{n} :=^{1} /_{n} .

Then

x_{n} < y_{n},

x_{n} < 0,

and

y_{n} > 0

for all

n .

However, these inequalities are not preserved by the limit operation as

lim_{n \to \infty} x_{n} = lim_{n \to \infty} y_{n} = 0 .

The moral of this example is that strict inequalities may become non-strict inequalities when limits are applied; if we know

x_{n} < y_{n}

for all

n,

we may only conclude

lim_{n \to \infty} x_{n} \leq lim_{n \to \infty} y_{n} .

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This issue is a common source of errors.

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Subsection 2.2.2 Continuity of algebraic operations

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Limits interact nicely with algebraic operations.

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Proposition 2.2.5.

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Let

{x_{n}}_{n = 1}^{\infty}

and

{y_{n}}_{n = 1}^{\infty}

be convergent sequences.

The sequence ${z_{n}}_{n = 1}^{\infty},$ where $z_{n} := x_{n} + y_{n},$ converges and

$lim_{n \to \infty} (x_{n} + y_{n}) = lim_{n \to \infty} z_{n} = lim_{n \to \infty} x_{n} + lim_{n \to \infty} y_{n} .$
The sequence ${z_{n}}_{n = 1}^{\infty},$ where $z_{n} := x_{n} - y_{n},$ converges and

$lim_{n \to \infty} (x_{n} - y_{n}) = lim_{n \to \infty} z_{n} = lim_{n \to \infty} x_{n} - lim_{n \to \infty} y_{n} .$
The sequence ${z_{n}}_{n = 1}^{\infty},$ where $z_{n} := x_{n} y_{n},$ converges and

$lim_{n \to \infty} (x_{n} y_{n}) = lim_{n \to \infty} z_{n} = (lim_{n \to \infty} x_{n}) (lim_{n \to \infty} y_{n}) .$
If $lim_{n \to \infty} y_{n} \neq 0$ and $y_{n} \neq 0$ for all $n \in N,$ then the sequence ${z_{n}}_{n = 1}^{\infty},$ where $z_{n} := \frac{x_{n}}{y_{n}},$ converges and

$lim_{n \to \infty} \frac{x_{n}}{y_{n}} = lim_{n \to \infty} z_{n} = \frac{lim_{n \to \infty} x_{n}}{lim_{n \to \infty} y_{n}} .$

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Proof.

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We start with i. Suppose

{x_{n}}_{n = 1}^{\infty}

and

{y_{n}}_{n = 1}^{\infty}

are convergent sequences and write

z_{n} := x_{n} + y_{n} .

Let

x := lim_{n \to \infty} x_{n},

y := lim_{n \to \infty} y_{n},

and

z := x + y .

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Let

ϵ > 0

be given. Find an

M_{1}

such that for all

n \geq M_{1},

we have

| x_{n} - x | <^{ϵ} /_{2} .

Find an

M_{2}

such that for all

n \geq M_{2},

we have

| y_{n} - y | <^{ϵ} /_{2} .

Take

M := max {M_{1}, M_{2}} .

For all

n \geq M,

we have

\begin{aligned} | z_{n} - z | & = | (x_{n} + y_{n}) - (x + y) | \\ = | x_{n} - x + y_{n} - y | \\ \leq | x_{n} - x | + | y_{n} - y | \\ < \frac{ϵ}{2} + \frac{ϵ}{2} = ϵ . \end{aligned}

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Therefore i is proved. Proof of ii is almost identical and is left as an exercise.

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Let us tackle iii. Suppose again that

{x_{n}}_{n = 1}^{\infty}

and

{y_{n}}_{n = 1}^{\infty}

are convergent sequences and write

z_{n} := x_{n} y_{n} .

Let

x := lim_{n \to \infty} x_{n},

y := lim_{n \to \infty} y_{n},

and

z := x y .

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Let

ϵ > 0

be given. Let

K := max {| x |, | y |,^{ϵ} /_{3}, 1} .

Find an

M_{1}

such that for all

n \geq M_{1},

we have

| x_{n} - x | < \frac{ϵ}{3 K} .

Find an

M_{2}

such that for all

n \geq M_{2},

we have

| y_{n} - y | < \frac{ϵ}{3 K} .

Take

M := max {M_{1}, M_{2}} .

For all

n \geq M,

we have

\begin{aligned} | z_{n} - z | & = | (x_{n} y_{n}) - (x y) | \\ = | (x_{n} - x + x) (y_{n} - y + y) - x y | \\ = | (x_{n} - x) y + x (y_{n} - y) + (x_{n} - x) (y_{n} - y) | \\ \leq | (x_{n} - x) y | + | x (y_{n} - y) | + | (x_{n} - x) (y_{n} - y) | \\ = | x_{n} - x | | y | + | x | | y_{n} - y | + | x_{n} - x | | y_{n} - y | \\ < \frac{ϵ}{3 K} K + K \frac{ϵ}{3 K} + \frac{ϵ}{3 K} \frac{ϵ}{3 K} (now notice that \frac{ϵ}{3 K} \leq 1 and K \geq 1) \\ \leq \frac{ϵ}{3} + \frac{ϵ}{3} + \frac{ϵ}{3} = ϵ . \end{aligned}

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Finally, we examine iv. Instead of proving iv directly, we prove the following simpler claim:

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Claim: If ${y_{n}}_{n = 1}^{\infty}$ is a convergent sequence such that $lim_{n \to \infty} y_{n} \neq 0$ and $y_{n} \neq 0$ for all $n \in N,$ then ${^{1} /_{y_{n}}}_{n = 1}^{\infty}$ converges and

lim_{n \to \infty} \frac{1}{y_{n}} = \frac{1}{lim_{n \to \infty} y_{n}} .

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Once the claim is proved, we take the sequence

{^{1} /_{y_{n}}}_{n = 1}^{\infty},

multiply it by the sequence

{x_{n}}_{n = 1}^{\infty}

and apply item iii.

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Proof of claim: Let

ϵ > 0

be given. Let

y := lim_{n \to \infty} y_{n} .

| y | \neq 0,

then

min {{| y |}^{2} \frac{ϵ}{2}, \frac{| y |}{2}} > 0 .

Find an

M

such that for all

n \geq M,

we have

| y_{n} - y | < min {{| y |}^{2} \frac{ϵ}{2}, \frac{| y |}{2}} .

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For all

n \geq M,

we have

| y - y_{n} | <^{| y |} /_{2},

and so

| y | = | y - y_{n} + y_{n} | \leq | y - y_{n} | + | y_{n} | < \frac{| y |}{2} + | y_{n} | .

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Subtracting

^{| y |} /_{2}

from both sides we obtain

^{| y |} /_{2} < | y_{n} |,

or in other words,

\frac{1}{| y_{n} |} < \frac{2}{| y |} .

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We finish the proof of the claim:

\begin{aligned} | \frac{1}{y_{n}} - \frac{1}{y} | & = | \frac{y - y_{n}}{y y_{n}} | \\ = \frac{| y - y_{n} |}{| y | | y_{n} |} \\ \leq \frac{| y - y_{n} |}{| y |} \frac{2}{| y |} \\ < \frac{{| y |}^{2} \frac{ϵ}{2}}{| y |} \frac{2}{| y |} = ϵ . \end{aligned}

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And we are done.

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By plugging in constant sequences, we get several easy corollaries. If

c \in R

and

{x_{n}}_{n = 1}^{\infty}

is a convergent sequence, then for example

lim_{n \to \infty} c x_{n} = c (lim_{n \to \infty} x_{n}) and lim_{n \to \infty} (c + x_{n}) = c + lim_{n \to \infty} x_{n} .

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Similarly, we find such equalities for constant subtraction and division.

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As we can take limits past multiplication we can show (exercise) that

lim_{n \to \infty} x_{n}^{k} = {(lim_{n \to \infty} x_{n})}^{k}

for all

k \in N .

That is, we can take limits past powers. Let us see if we can do the same with roots.

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Proposition 2.2.6.

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Let

{x_{n}}_{n = 1}^{\infty}

be a convergent sequence such that

x_{n} \geq 0

for all

n \in N .

Then

lim_{n \to \infty} \sqrt{x_{n}} = \sqrt{lim_{n \to \infty} x_{n}} .

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Of course, to even make this statement, we need to apply Corollary 2.2.4 to show that

lim_{n \to \infty} x_{n} \geq 0,

so that we can take the square root without worry.

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Proof.

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Let

{x_{n}}_{n = 1}^{\infty}

be a convergent sequence and let

x := lim_{n \to \infty} x_{n} .

As we just mentioned,

x \geq 0 .

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First suppose

x = 0 .

Let

ϵ > 0

be given. Then there is an

M

such that for all

n \geq M,

we have

x_{n} = | x_{n} | < ϵ^{2},

or in other words,

\sqrt{x_{n}} < ϵ .

Hence,

| \sqrt{x_{n}} - \sqrt{x} | = \sqrt{x_{n}} < ϵ .

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Now suppose

x > 0

(and hence

\sqrt{x} > 0

\begin{aligned} | \sqrt{x_{n}} - \sqrt{x} | & = | \frac{x_{n} - x}{\sqrt{x_{n}} + \sqrt{x}} | \\ = \frac{1}{\sqrt{x_{n}} + \sqrt{x}} | x_{n} - x | \\ \leq \frac{1}{\sqrt{x}} | x_{n} - x | . \end{aligned}

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We leave the rest of the proof to the reader.

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A similar proof works for the

k

th root. That is, we also obtain

lim_{n \to \infty} x_{n}^{1 / k} = {(lim_{n \to \infty} x_{n})}^{1 / k} .

We leave this to the reader as a challenging exercise.

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We may also want to take the limit past the absolute value sign. The converse of this proposition is not true, see Exercise 2.1.7 part b).

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Proposition 2.2.7.

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{x_{n}}_{n = 1}^{\infty}

is a convergent sequence, then

{| x_{n} |}_{n = 1}^{\infty}

is convergent and

lim_{n \to \infty} | x_{n} | = | lim_{n \to \infty} x_{n} | .

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Proof.

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We simply note the reverse triangle inequality

| | x_{n} | - | x | | \leq | x_{n} - x | .

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Hence if

| x_{n} - x |

can be made arbitrarily small, so can

| | x_{n} | - | x | | .

Details are left to the reader.

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Let us see an example putting the propositions above together. Since

lim_{n \to \infty}^{1} /_{n} = 0,

then

lim_{n \to \infty} | \sqrt{1 +^{1} /_{n}} -^{100} /_{n^{2}} | = | \sqrt{1 + (lim_{n \to \infty}^{1} /_{n})} - 100 (lim_{n \to \infty}^{1} /_{n}) (lim_{n \to \infty}^{1} /_{n}) | = 1.

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That is, the limit on the left-hand side exists because the right-hand side exists. You really should read the equality above from right to left.

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On the other hand you must apply the propositions carefully. For example, by rewriting the expression with common denominator first we find

lim_{n \to \infty} (\frac{n^{2}}{n + 1} - n) = - 1 .

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However,

{\frac{n^{2}}{n + 1}}_{n = 1}^{\infty}

and

{n}_{n = 1}^{\infty}

are not convergent, so

(lim_{n \to \infty} \frac{n^{2}}{n + 1}) - (lim_{n \to \infty} n)

is nonsense.

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Subsection 2.2.3 Recursively defined sequences

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Now that we know we can interchange limits and algebraic operations, we can compute the limits of many sequences. One such class are recursively defined sequences, that is, sequences where the next number in the sequence is computed using a formula from a fixed number of preceding elements in the sequence.

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Example 2.2.8.

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Let

{x_{n}}_{n = 1}^{\infty}

be defined by

x_{1} := 2

and

x_{n + 1} := x_{n} - \frac{x_{n}^{2} - 2}{2 x_{n}} .

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We must first find out if this sequence is well-defined; we must show we never divide by zero. Then we must find out if the sequence converges. Only then can we attempt to find the limit.

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So let us prove that for all

n

x_{n}

exists and

x_{n} > 0

(so the sequence is well-defined and bounded below). Let us show this by induction. We know that

x_{1} = 2 > 0 .

For the induction step, suppose

x_{n}

exists and

x_{n} > 0 .

Then

x_{n + 1} = x_{n} - \frac{x_{n}^{2} - 2}{2 x_{n}} = \frac{2 x_{n}^{2} - x_{n}^{2} + 2}{2 x_{n}} = \frac{x_{n}^{2} + 2}{2 x_{n}} .

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It is always true that

x_{n}^{2} + 2 > 0,

and as

x_{n} > 0,

then

x_{n + 1} = \frac{x_{n}^{2} + 2}{2 x_{n}} > 0 .

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Next let us show that the sequence is monotone decreasing. If we show that

x_{n}^{2} - 2 \geq 0

for all

n,

then

x_{n + 1} \leq x_{n}

for all

n .

Obviously

x_{1}^{2} - 2 = 4 - 2 = 2 > 0 .

For an arbitrary

n,

we have

x_{n + 1}^{2} - 2 = {(\frac{x_{n}^{2} + 2}{2 x_{n}})}^{2} - 2 = \frac{x_{n}^{4} + 4 x_{n}^{2} + 4 - 8 x_{n}^{2}}{4 x_{n}^{2}} = \frac{x_{n}^{4} - 4 x_{n}^{2} + 4}{4 x_{n}^{2}} = \frac{{(x_{n}^{2} - 2)}^{2}}{4 x_{n}^{2}} .

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Since squares are nonnegative,

x_{n + 1}^{2} - 2 \geq 0

for all

n .

Therefore,

{x_{n}}_{n = 1}^{\infty}

is monotone decreasing and bounded (

x_{n} > 0

for all

n

), and so the limit exists. It remains to find the limit.

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Write

2 x_{n} x_{n + 1} = x_{n}^{2} + 2 .

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Since

{x_{n + 1}}_{n = 1}^{\infty}

is the 1-tail of

{x_{n}}_{n = 1}^{\infty},

it converges to the same limit. Let us define

x := lim_{n \to \infty} x_{n} .

Take the limit of both sides to obtain

2 x^{2} = x^{2} + 2,

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x^{2} = 2 .

x_{n} > 0

for all

n

we get

x \geq 0,

and therefore

x = \sqrt{2} .

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You may have seen the sequence above before. It is Newton’s method

Named after the English physicist and mathematician Isaac Newton (1642–1726/7).

for finding the square root of 2. This method comes up often in practice and converges very rapidly. We used the fact that

x_{1}^{2} - 2 > 0,

although it was not strictly needed to show convergence by considering a tail of the sequence. The sequence converges as long as

x_{1} \neq 0,

although with a negative

x_{1}

we would arrive at

x = - \sqrt{2} .

By replacing the 2 in the numerator we obtain the square root of any positive number. These statements are left as an exercise.

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You should, however, be careful. Before taking any limits, you must make sure the sequence converges. Let us see an example.

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Example 2.2.9.

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Suppose

x_{1} := 1

and

x_{n + 1} := x_{n}^{2} + x_{n} .

If we blindly assumed that the limit exists (call it

x

), then we would get the equation

x = x^{2} + x,

from which we might conclude

x = 0 .

However, it is not hard to show that

{x_{n}}_{n = 1}^{\infty}

is unbounded and therefore does not converge.

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The thing to notice in this example is that the method still works, but it depends on the initial value

x_{1} .

If we set

x_{1} := 0,

then the sequence converges and the limit really is 0. An entire branch of mathematics, called dynamics, deals precisely with these issues. See Exercise 2.2.14.

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Subsection 2.2.4 Some convergence tests

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It is not always necessary to go back to the definition of convergence to prove that a sequence is convergent. We first give a simple convergence test. The main idea is that

{x_{n}}_{n = 1}^{\infty}

converges to

x

if and only if

{| x_{n} - x |}_{n = 1}^{\infty}

converges to zero.

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Proposition 2.2.10.

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Let

{x_{n}}_{n = 1}^{\infty}

be a sequence. Suppose there is an

x \in R

and a convergent sequence

{a_{n}}_{n = 1}^{\infty}

such that

lim_{n \to \infty} a_{n} = 0

🔗

and

| x_{n} - x | \leq a_{n} for all n \in N .

🔗

Then

{x_{n}}_{n = 1}^{\infty}

converges and

lim_{n \to \infty} x_{n} = x .

🔗

Proof.

🔗

Let

ϵ > 0

be given. Note that

a_{n} \geq 0

for all

n .

Find an

M \in N

such that for all

n \geq M,

we have

a_{n} = | a_{n} - 0 | < ϵ .

Then, for all

n \geq M,

we have

| x_{n} - x | \leq a_{n} < ϵ .

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As the proposition shows, to study when a sequence has a limit is the same as studying when another sequence goes to zero. In general, it may be hard to decide if a sequence converges, but for certain sequences there exist easy to apply tests that tell us if the sequence converges or not. Let us see one such test. First, let us compute the limit of a certain specific sequence.

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Proposition 2.2.11.

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Let

c > 0 .

If $c < 1,$ then

$lim_{n \to \infty} c^{n} = 0.$
If $c > 1,$ then ${c^{n}}_{n = 1}^{\infty}$ is unbounded.

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Proof.

🔗

First consider

c < 1 .

c > 0,

then

c^{n} > 0

for all

n \in N

by induction. As

c < 1,

then

c^{n + 1} < c^{n}

for all

n .

{c^{n}}_{n = 1}^{\infty}

is a decreasing sequence that is bounded below. Hence, it is convergent. Let

x := lim_{n \to \infty} c^{n} .

The 1-tail

{c^{n + 1}}_{n = 1}^{\infty}

also converges to

x .

Taking the limit of both sides of

c^{n + 1} = c \cdot c^{n},

we obtain

x = c x,

(1 - c) x = 0 .

It follows that

x = 0

c \neq 1 .

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Now consider

c > 1 .

Let

B > 0

be arbitrary. As

^{1} /_{c} < 1,

then

{{(^{1} /_{c})}^{n}}_{n = 1}^{\infty}

converges to

0 .

Hence for some large enough

n,

we get

\frac{1}{c^{n}} = {(\frac{1}{c})}^{n} < \frac{1}{B} .

🔗

In other words,

c^{n} > B,

and

B

is not an upper bound for

{c^{n}}_{n = 1}^{\infty} .

B

was arbitrary,

{c^{n}}_{n = 1}^{\infty}

is unbounded.

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In the proposition above, the ratio of the

(n + 1)

th term and the

n

th term is

c .

We generalize this simple result to a larger class of sequences. The following lemma will come up again once we get to series.

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Lemma 2.2.12. Ratio test for sequences.

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Let

{x_{n}}_{n = 1}^{\infty}

be a sequence such that

x_{n} \neq 0

for all

n

and such that the limit

L := lim_{n \to \infty} \frac{| x_{n + 1} |}{| x_{n} |} exists.

If $L < 1,$ then ${x_{n}}_{n = 1}^{\infty}$ converges and $lim_{n \to \infty} x_{n} = 0 .$
If $L > 1,$ then ${x_{n}}_{n = 1}^{\infty}$ is unbounded (hence diverges).

🔗

L

exists, but

L = 1,

the lemma says nothing. We cannot make any conclusion based on that information alone. For example, the sequence

{^{1} /_{n}}_{n = 1}^{\infty}

converges to zero, but

L = 1 .

The constant sequence

{1}_{n = 1}^{\infty}

converges to 1, not zero, and

L = 1 .

The sequence

{{(- 1)}^{n}}_{n = 1}^{\infty}

does not converge at all, and

L = 1

as well. Finally, the sequence

{n}_{n = 1}^{\infty}

is unbounded, yet again

L = 1 .

The statement of the lemma may be strengthened somewhat, see exercises 2.2.13 and 2.3.15.

🔗

Proof.

🔗

Suppose

L < 1 .

\frac{| x_{n + 1} |}{| x_{n} |} \geq 0

for all

n,

then

L \geq 0 .

Pick

r

such that

L < r < 1 .

We wish to compare the sequence

{x_{n}}_{n = 1}^{\infty}

to the sequence

{r^{n}}_{n = 1}^{\infty} .

The idea is that while the ratio

\frac{| x_{n + 1} |}{| x_{n} |}

is not going to be less than

L

eventually, it will eventually be less than

r,

which is still less than 1. The intuitive idea of the proof is illustrated in Figure 2.4.

Figure 2.4. The short lines represent the ratios

\frac{| x_{n + 1} |}{| x_{n} |}

approaching

L < 1 .

🔗

r - L > 0,

there exists an

M \in N

such that for all

n \geq M,

we have

| \frac{| x_{n + 1} |}{| x_{n} |} - L | < r - L .

🔗

Therefore, for

n \geq M,

\frac{| x_{n + 1} |}{| x_{n} |} - L < r - L or \frac{| x_{n + 1} |}{| x_{n} |} < r .

🔗

For

n > M

(that is for

n \geq M + 1

) write

| x_{n} | = | x_{M} | \frac{| x_{M + 1} |}{| x_{M} |} \frac{| x_{M + 2} |}{| x_{M + 1} |} \dots \frac{| x_{n} |}{| x_{n - 1} |} < | x_{M} | r r \dots r = | x_{M} | r^{n - M} = (| x_{M} | r^{- M}) r^{n} .

🔗

The sequence

{r^{n}}_{n = 1}^{\infty}

converges to zero and hence

| x_{M} | r^{- M} r^{n}

converges to zero. By Proposition 2.2.10, the

M

-tail

{x_{n}}_{n = M + 1}^{\infty}

converges to zero and therefore

{x_{n}}_{n = 1}^{\infty}

converges to zero.

🔗

Now suppose

L > 1 .

Pick

r

such that

1 < r < L .

L - r > 0,

there exists an

M \in N

such that for all

n \geq M

| \frac{| x_{n + 1} |}{| x_{n} |} - L | < L - r .

🔗

Therefore,

\frac{| x_{n + 1} |}{| x_{n} |} > r .

🔗

Again for

n > M,

write

| x_{n} | = | x_{M} | \frac{| x_{M + 1} |}{| x_{M} |} \frac{| x_{M + 2} |}{| x_{M + 1} |} \dots \frac{| x_{n} |}{| x_{n - 1} |} > | x_{M} | r r \dots r = | x_{M} | r^{n - M} = (| x_{M} | r^{- M}) r^{n} .

🔗

The sequence

{r^{n}}_{n = 1}^{\infty}

is unbounded (since

r > 1

), and so

{x_{n}}_{n = 1}^{\infty}

cannot be bounded (if

| x_{n} | \leq B

for all

n,

then

r^{n} < \frac{B}{| x_{M} |} r^{M}

for all

n > M,

which is impossible). Consequently,

{x_{n}}_{n = 1}^{\infty}

cannot converge.

🔗

Example 2.2.13.

🔗

A simple application of the lemma above is to prove

lim_{n \to \infty} \frac{2^{n}}{n!} = 0 .

🔗

Proof: Compute

\frac{2^{n + 1} / (n + 1)!}{2^{n} / n!} = \frac{2^{n + 1}}{2^{n}} \frac{n!}{(n + 1)!} = \frac{2}{n + 1} .

🔗

It is not hard to see that

{\frac{2}{n + 1}}_{n = 1}^{\infty}

converges to zero. The conclusion follows by the lemma.

🔗

Example 2.2.14.

🔗

A more complicated (and useful) application of the ratio test is to prove

lim_{n \to \infty} n^{1 / n} = 1 .

🔗

Proof: Let

ϵ > 0

be given. Consider the sequence

{\frac{n}{{(1 + ϵ)}^{n}}}_{n = 1}^{\infty} .

Compute

\frac{(n + 1) / {(1 + ϵ)}^{n + 1}}{n / {(1 + ϵ)}^{n}} = \frac{n + 1}{n} \frac{1}{1 + ϵ} .

🔗

The limit of

\frac{n + 1}{n} = 1 + \frac{1}{n}

n \to \infty

is 1, and so

lim_{n \to \infty} \frac{(n + 1) / {(1 + ϵ)}^{n + 1}}{n / {(1 + ϵ)}^{n}} = \frac{1}{1 + ϵ} < 1 .

🔗

Therefore,

{\frac{n}{{(1 + ϵ)}^{n}}}_{n = 1}^{\infty}

converges to 0. In particular, there exists an

M

such that for

n \geq M,

we have

\frac{n}{{(1 + ϵ)}^{n}} < 1,

n < {(1 + ϵ)}^{n},

n^{1 / n} < 1 + ϵ .

n \geq 1,

then

n^{1 / n} \geq 1,

and so

0 \leq n^{1 / n} - 1 < ϵ .

Consequently,

lim_{n \to \infty} n^{1 / n} = 1 .

🔗

Exercises 2.2.5 Exercises

🔗

2.2.1.

🔗

Prove Corollary 2.2.4. Hint: Use constant sequences and Lemma 2.2.3.

🔗

2.2.2.

🔗

Prove part ii of Proposition 2.2.5.

🔗

2.2.3.

🔗

Prove that if

{x_{n}}_{n = 1}^{\infty}

is a convergent sequence,

k \in N,

then

lim_{n \to \infty} x_{n}^{k} = {(lim_{n \to \infty} x_{n})}^{k} .

🔗

Hint: Use induction.

🔗

2.2.4.

🔗

Suppose

x_{1} := \frac{1}{2}

and

x_{n + 1} := x_{n}^{2} .

Show that

{x_{n}}_{n = 1}^{\infty}

converges and find

lim_{n \to \infty} x_{n} .

Hint: You cannot divide by zero!

🔗

2.2.5.

🔗

Let

x_{n} := \frac{n - \cos (n)}{n} .

Use the squeeze lemma to show that

{x_{n}}_{n = 1}^{\infty}

converges and find the limit.

🔗

2.2.6.

🔗

Let

x_{n} := \frac{1}{n^{2}}

and

y_{n} := \frac{1}{n} .

Define

z_{n} := \frac{x_{n}}{y_{n}}

and

w_{n} := \frac{y_{n}}{x_{n}} .

{z_{n}}_{n = 1}^{\infty}

and

{w_{n}}_{n = 1}^{\infty}

converge? What are the limits? Can you apply Proposition 2.2.5? Why or why not?

🔗

2.2.7.

🔗

True or false, prove or find a counterexample. If

{x_{n}}_{n = 1}^{\infty}

is a sequence such that

{x_{n}^{2}}_{n = 1}^{\infty}

converges, then

{x_{n}}_{n = 1}^{\infty}

converges.

🔗

2.2.8.

🔗

Show that

lim_{n \to \infty} \frac{n^{2}}{2^{n}} = 0 .

🔗

2.2.9.

🔗

Suppose

{x_{n}}_{n = 1}^{\infty}

is a sequence,

x \in R,

and

x_{n} \neq x

for all

n \in N .

Suppose the limit

L := lim_{n \to \infty} \frac{| x_{n + 1} - x |}{| x_{n} - x |}

🔗

exists and

L < 1 .

Show that

{x_{n}}_{n = 1}^{\infty}

converges to

x .

🔗

2.2.10.

🔗

(Challenging) Let

{x_{n}}_{n = 1}^{\infty}

be a convergent sequence such that

x_{n} \geq 0

and

k \in N .

Then

lim_{n \to \infty} x_{n}^{1 / k} = {(lim_{n \to \infty} x_{n})}^{1 / k} .

🔗

Hint: Find an expression

q

such that

\frac{x_{n}^{1 / k} - x^{1 / k}}{x_{n} - x} = \frac{1}{q} .

🔗

2.2.11.

🔗

Let

r > 0 .

Show that starting with an arbitrary

x_{1} \neq 0,

the sequence defined by

x_{n + 1} := x_{n} - \frac{x_{n}^{2} - r}{2 x_{n}}

🔗

converges to

\sqrt{r}

x_{1} > 0

and

- \sqrt{r}

x_{1} < 0 .

🔗

2.2.12.

🔗

Let

{a_{n}}_{n = 1}^{\infty}

and

{b_{n}}_{n = 1}^{\infty}

be sequences.

Suppose ${a_{n}}_{n = 1}^{\infty}$ is bounded and ${b_{n}}_{n = 1}^{\infty}$ converges to 0. Show that ${a_{n} b_{n}}_{n = 1}^{\infty}$ converges to 0.
Find an example where ${a_{n}}_{n = 1}^{\infty}$ is unbounded, ${b_{n}}_{n = 1}^{\infty}$ converges to 0, and ${a_{n} b_{n}}_{n = 1}^{\infty}$ is not convergent.
Find an example where ${a_{n}}_{n = 1}^{\infty}$ is bounded, ${b_{n}}_{n = 1}^{\infty}$ converges to some $x \neq 0,$ and ${a_{n} b_{n}}_{n = 1}^{\infty}$ is not convergent.

🔗

2.2.13.

🔗

(Easy) Prove the following stronger version of Lemma 2.2.12, the ratio test. Suppose

{x_{n}}_{n = 1}^{\infty}

is a sequence such that

x_{n} \neq 0

for all

n .

Prove that if there exists an $r < 1$ and $M \in N$ such that

$\frac{| x_{n + 1} |}{| x_{n} |} \leq r for all n \geq M,$

then ${x_{n}}_{n = 1}^{\infty}$ converges to $0 .$
Prove that if there exists an $r > 1$ and $M \in N$ such that

$\frac{| x_{n + 1} |}{| x_{n} |} \geq r for all n \geq M,$

then ${x_{n}}_{n = 1}^{\infty}$ is unbounded.