This is the sequence of values of the function

Sequences of numbers appear in diverse situations. If you divide a cake in half, and then divide the remaining half in half, and continue dividing in half indefinitely, then the fraction of cake remaining at each step forms the sequence

This is the sequence of values of the function

1 1 1 1, , , ...2 2 4 8nf n

1n

Formally, a sequence is an ordered collection of numbers defined by a function f (n) on a set of integers. The values an = f (n) are called the terms of the sequence, and n is called the index. Informally, we think of a sequence {an} as a list of terms:

1 2 3 4, , , , ...a a a aThe sequence does not have to start at n = 1, but may start at n = 0, n = 2, or any other integer. When an is given by a formula, we refer to an as the general term.

11nan

1 nna n

2

2

364.54nnb

n

Calculator!

Compute a2, a3, a4 for the sequence defined recursively by

1 11

1 21, 2n n

n

a a aa

The sequence in the next example is defined recursively. The first term is given and the nth term an is computed in terms of the preceding term an-1.

2

3

4

1 21 1.52 11 3 2 1.41672 2 3 / 21 17 2 1.414216

32

1712

5774082 12 17 /12

a

a

a

Decimal to Fraction

Our main goal is to study convergence of sequences. A sequence {an} converges to a limit L if |an − L| becomes arbitrary small when n is sufficiently large. Here is the formal definition.DEFINITION Limit of a Sequence We say that {an} converges to a limit L, and we write

if, for every ε > 0, there is a number M such that |an − L| < ε for all n > M.

• If no limit exists, we say that {an} diverges.• If the terms increase without bound, we say that {an} diverges to infinity.

lim or n nna L a L

Proof

If {an} converges, then its limit L is unique. A good way to visualize the limit is to plot the points (1, a1), (2, a2), (3, a3),…. The sequence converges to L if, for every ε > 0, the plotted points eventually remain within an ε-band around the horizontal line y = L.

11, a

The figure below shows the plot of a sequence converging to L = 1.

11,a

On the other hand, we can show that the sequence an = cos n has no limit.

11, a

4Let . Prove formally that lim 1.1n nn

na an

Show that lim 1 with a numerical and graphical investigation.nn

a

lim 1nna

lim 1nna

Leading Terms!4lim lim lim1 11n n n

n nn n

3 31 1n n

4 4 1 31 11 1 1 1n

n n nan n n n

4Let . Prove formally that lim 1.1n nn

na an

Solution The definition requires us to find, for every ε > 0, a number M such that

1 , na n M

3 11na

n

3In other words, is valid with 11 ,

!

.na n M

E

M

Q D

Solve for :n

Formal Definition of Limit

So given any , we can find the number ofterms that it will take for the sequence

to be within units of the limit.n

n

a

Note the following two facts about sequences:

• The limit does not change if we change or drop finitely many terms of the sequence.

• If C is a constant and an = C for all n sufficiently large, thenlim .nn

a C

Many of the sequences we consider are defined by functions; that is, an = f (n) for some function f (x). For example,

A fact we will use often is that if f (x) approaches a limit L as x → ∞, then the sequence an = f (n) approaches the same limit L. Indeed, for all ε > 0, we can find M so that |f (n) − L| < ε for all x > M. It follows automatically that |f (n) − L| < ε for all integers n > M.

1 1 is defined by nn xa f x

n x

If f (x) converges to L, then the sequence an = f (n) also converges to L.

THEOREM 1 Sequence Defined by a Function

lim exists converges to the same limit:nxf x a f n

Find the limit of the sequence

lim limnn xa f x

2 2 2 2

2 2 2 2

2 2 3 2 4 2 5 2, , , , ...2 3 4 5

2

2 2

2 21nna

n n

2

2lim li 1 1mnn xa

x

2

lnCalculate lim .n

n nn

THEOREM 1 Sequence Defined by a Function

lim exists converges to the same limit:nxf x a f n

lim limnn xa f x

2 2

ln ln 1 1/lim lim li 0m2n x x

n n x x xn x x

Indeterminant Form ' ' L Hopital s Rule

Direct Substitution yields...

The limit of the Balmer wavelengths bn in the next example plays a role in physics and chemistry because it determines the ionization energy of the hydrogen atom. Table 1 suggests that bn approaches 364.5. Figure 6 shows the graph, and in Figure 7, the wavelengths are shown “crowding in” toward their limiting value.

Calculate the limit of the Balmer wavelengths 2

2

364.5 , where 3.4nnb n

n

2

2

364.5 364.lim lim4

5nn x

xbx

...Leading Terms

lim limnn xa f x

A geometric sequence is a sequence an = crn, where c and r are nonzero constants. Each term is r times the previous term; that is, an/an-1 = r. The number r is called the common ratio. For instance, if r = 3 and c = 2, we obtain the sequence (starting at n = 0)

In the next example, we determine when a geometric series converges. Recall that {an} diverges to ∞ if the terms an increase beyond all bounds; that is,

for every number N, an > N for all sufficiently large nlim if,nna

We define lim

similarly, with 0.nn

a

c

If r > 1, the geometric sequence an = rn diverges to ∞.

2 3 4 52, 2 3, 2 3 , 2 3 , 2 3 , 2 3 ,

Geometric Sequences with r ≥ 0 Prove that for r ≥ 0 and c > 0,

f (x) = crx. If 0 ≤ r < 1, then

If r > 1, then both f (x) and the sequence {crn} diverge to ∞ (because c > 0).

If r = 1, then crn = c for all n, and the limit is c.

lim lim lim 0n x x

n x xcr cr c r

Function

Laws of Limits

THEOREM 2 Limit Laws for Sequences Assume that {an} and {bn} are convergent sequences with

Then:

i lim lim lim

ii lim lim lim

limiii lim , 0

lim

iv lim lim

n n n nn n n

n n n nn n n

nn n

nn nn

n nn n

a b a b L M

a b a b LM

aa L Mb b M

ca c a cL

lim , limn nn na L b M

THEOREM 3 Squeeze Theorem for Sequences Let {an}, {bn}, {cn} be sequences such that for some number M,

Then

Show that if lim 0, then lim 0.n nn na a

for and lim limn n n n nn nb a c n M b c L

lim nna L

n n na a a

lim 0 lim lim 0n n nn n na a a

by the squeeze THM lim 0nna

2 2 2

Laws of Limits

Geometric Sequences with r < 0 Prove that for c 0.

If -1 < r < 0, then

lim lim lim 0n x x

n x xcr cr c r

If r = -1, then

1 , which diverges (oscilates between & )nncr c c c If r < -1, then

alternates sign and grows in magnitude (diverges)ncr

0, 1 0diverges, 1lim {n r

rncr

As another application of the Squeeze Theorem, consider the sequence

Both the numerator and the denominator grow without bound, so it is not clear in advance whether {an} converges. The graph & table suggest that an increases initially and then tends to zero. In the next example, we verify that an = Rn/n! converges to zero for all R. This fact is used in the discussion of Taylor series in Section 10.7.

5!

n

nan

Let M be a positive integer such that

Since CR/n → 0, the Squeeze Theorem gives us

The first M factors are ≥ 1 and the last n − M factors are < 1. If we lump together the first M factors and call the product C, and drop all the remaining factors except the last factor R/n, we see that

Prove that lim 0 for all 0.!

n

n

R Rn

, na L n M

1M R M

For n > M, we write Rn/n! as a product of n factors:

0!

nR CRn n

as claimed. If R < 0, the limit is also zero by Example 7 because |Rn/n!| tends to zero.

1RM M

. . 4, in i e n n

lim / ! 0n

nR n

You'll see why, with ournext compound inequality.

What if in ?R

Given a sequence {an} and a function f (x), we can form the new sequence f (an). It is useful to know that if f (x) is continuous and an → L, then f (an) → f (L).

THEOREM 4 If f (x) is continuous and lim , thennna L

In other words, we may “bring a limit inside a continuous function.”

lim limn nn nf a f a f L

3lim lim1

3nn n

nL an

Apply Theorem 4 to the sequence

and to the functions (a) f (x) = ex and (b) g (x) = x2.

With f (x) = ex we have 3

1.n

na n

nf a e e According to Theorem 4,

3lim

33

11lim lim lim n

nnnn

n nn n ne ef a f a e

With g(x) = x2 we have 2n ng a a , and according to Theorem 4,

2 2

23 3lim lim lim lim 931 1n nn n n n

n ng a g an n

31n

nan

Of great importance for understanding convergence are the concepts of a bounded sequence and a monotonic sequence.

DEFINITION Bounded Sequences A sequence {an} is:• Bounded from above if there is a number M such that an ≤ M for

all n. The number M is called an upper bound.• Bounded from below if there is a number m such that an ≥ m for

all n. The number m is called a lower bound.The sequence {an} is called bounded if it is bounded from above and below. A sequence that is not bounded is called an unbounded sequence.

Upper and lower bounds are not unique. If M is an upper bound, then any larger number is also an upper bound, and if m is a lower bound, then any smaller number is also a lower bound (Figure 11).

As we might expect, a convergent sequence {an} is necessarily bounded because the terms an get closer and closer to the limit. This fact is recorded in the next theorem.

THEOREM 5 Convergent Sequences Are Bounded If {an} converges, then {an} is bounded.

There are two ways that a sequence {an} can diverge. One way is by being unbounded. For example, the unbounded sequence an = n diverges:

1, 2, 3, 4, 5, 6,…

However, a sequence can diverge even if it is bounded. This is the case with an = (−1)n+1, whose terms an bounce back and forth but never settle down to approach a limit:

1, −1, 1, −1, 1, −1,…

There is no surefire method for determining whether a sequence {an} converges, unless the sequence happens to be both bounded and monotonic. By definition, {an} is monotonic if it is either increasing or decreasing:

• {an} is increasing if an < an+1 for all n.• {an} is decreasing if an > an+1 for all n.

na

For 1, in ...n n is bounded and divergesna

Intuitively, if {an} is increasing and bounded above by M, then the terms must bunch up near some limiting value L that is not greater than M (Figure 12).

THEOREM 6 Bounded Monotonic Sequences Converge • If {an} is increasing and an ≤ M, then {an} converges and • If {an} is decreasing and an ≥ m, then {an} converges and

lim .nna M

lim .nna M

Verify that 1 is decreasing and bounded below.Does lim exist?

n

nn

a n na

1/ 2 1/ 2

Let 1

1 1' 12 2

1 1 0, 02 1 2

f x x x

f x x x

xx x

is decreasingna f n

Also, 0 na n

the sequence has a lower bound 0m

By THM 6, lim 0.nna L

Find lim 1 analytically.x

x x

1lim 1 lim 11

1l m1

0i

x x

x

x xx x x xx x

x x

2

2 2 2 2... 2 2 2 2... 2

2 2,00, because is positive & increasing 2n

L L

L L lL a L

Show that the following sequence is bounded and increasing:

Prove that lim exists and compute its value.nnL a

Step 1. Show that {an} is bounded above.We claim that M = 2 is an upper bound. We certainly have a1 < 2 because 1 2 1.414.a On the other hand, 12 2.n na a This is true because

1 2 2 2 2.n na a

Now, since a1 < 2, we can apply to conclude that a2 < 2. Similarly, a2 < 2 implies a3 < 2, and so on for all n. Formally speaking, this is a proof by induction.

1 2 2 2 2n na a

Step 2. Show that {an} is increasing.Since an is positive and an < 2, we have 1 2 .n n n n na a a a a This shows that {an} is increasing.We conclude that the limit L exists and hence L = 2.