Infinite Sequences and Series



a
a
a
a
a
a

a

A sequence is an ordered set of numbers 
,
n
1
2
3
4
5
n
an = L (and L is finite),
where an denotes the nth term. The sequence converges if lim
n 
otherwise it diverges.
an = ∞ we say that the limit exists, but the sequence diverges.
Note: if lim
n 
an = lim
cn = L, then
Theorem 1) Sandwich Theorem: if an < bn < cn, and lim
n 
n 
lim
bn = L also.
n 
| an | = 0, if, and only if lim
an = 0.
Theorem 2) lim
n
n 
f (x) = L and f(n) = an, then
Theorem 3) If f (x) is defined for real numbers x, lim
x
lim
an = L also. (This allows the use of l'Hopital's rule to find the limits of sequences.)
n 
x
Theorem 4) A geometric sequence {xn} has limit zero for |x| < 1, but lim
n
n
does not
n
1 = 1 when x = 1
exist for |x| > 1 and x = -1. lim
n 
A sequence is increasing if for all values of n, an < an+1; a sequence is decreasing
if for all values of n, an > an+1. A monotonic sequence is either increasing or decreasing.
A sequence is bounded above if there exists a number M such that M > an and bounded
below if there is a number m such that m < an . A bounded sequence is bounded both
above and below.
Theorem 5) A bounded monotonic sequence converges.
Given a sequence {ak} , if we add the terms we get a series. An infinite series is denoted

by:
a
k
k 1

Given a series
a
k 1
k
, we can define the partial sums
n
Sn ak These form a new sequence {Sn}. If {Sn} converges to S (that is
k1
lim S n = S) then we say the series converges and S is called the sum of the series.
n 
Otherwise the series diverges. Although it is sometimes possible to find the exact value of
a series, in general this is a very difficult problem. The first step with a series is to
determine whether it converges or diverges. If it does converge, the sum can usually be
estimated numerically.

Geometric Series:
x
k 0
k
converges if |x| < 1 and diverges otherwise.
The sum of this series is 1/(1 - x).

n
1
1
Harmonic Series:  The partial sums are usually denoted by Hn  Since Hn
k 0 k
k1 k
grows like ln(n), this series diverges.

a
Theorem 6) If
ak = 0.
converges, then lim
k 
k
k 1

Theorem 7) If  ak and
k 1




 bk both converge, then
a b converges also and
k 1

k
k1
k


a

b
a

bAlso for any constant c, 
ca c
a


k

1
k
k
k

1
k
k
k

1
k
k
1
k
1
k
Tests for Convergence

a
ak  0 then
Nth Term Divergence Test If lim
k
diverges.
k
k 1

a
ak = 0 the series
The converse of this is false. Just because the lim
k 
k 1
may
k
not converge. Write an example of this here: ____________
Integral Test: Let f be a positive, decreasing and continuous function for x > b.

a
If ak = f(k) then

whenever

b

converges whenever
k
k 1

b

f (x)dx converges and
a
k 1
k
diverges
f (x)dx diverges.

P - series Test:
1
k
k 1
converges for p > 1 and diverges for p < 1.
p

Comparison Test: If 0 < an < bn then: if
b
k 1
k

converges so does
a
k 1
k

; if
a
k 1
k

diverges so does
b
k 1
k
a bk = c and c is positive and finite
Limit Comparison Test: If 0 < ak , 0 < bk, lim
k k
then:

 bk
k 1

and
a
k 1
k
either both converge or they both diverge.
a bk = 0 and
If lim
k k
a bk =  and
If lim
k k

 bk converges, so does
k 1

 bk diverges, so does
k 1

a
k
k 1

a
k 1
k
Common series used for comparisons are geometric series and p-series.
If 0 < ak, then

k

1
k1
2
3
4
5
6
k

1
(

1
)
a

a

a

a

a

a

a



(

1
)
a


is

n

1
n
called an alternating series.
Alternating Series Test: (Leibniz’ Test)
ak = 0 then
If {ak} is decreasing and lim
k 

A series  ak is called absolutely convergent if
k 1

A series
 ak may converge while
k 1

(1)
k1
k1
ak converges.

a
k 1
k
converges.

a
k 1
k
diverges. Example? _________

 ak implies convergence of
Theorem 8) Convergence of
k 1

a
k 1
k
. This means that all of
the tests above can be applied to { |an| } if {an} contains both positive and negative
terms.

an1
Ratio Test: Given a series  ak , let lnim
If  < 1 then

 a
k 1
n

a
k 1
converges
k

absolutely; if  > 1 then
a
k 1
k
diverges and if  = 1 no conclusion can be drawn.

Root Test: Given a series
a
k 1
k
m | an | If  < 1 then
, let  lni

n

a
k 1
k
converges

absolutely; if  > 1 then
a
k 1
k
diverges and  = 1 no conclusion can be drawn.
These 2 different formulas for  both give the same value, provided the limits exist.
Power Series

The function f(x) =
a (xc)
k
k0
k
is called a power series centered at c or about c.
Of course, this function only makes sense at those points x where the series converges.
Power series, if they converge at more than one point, converge absolutely on an open
interval centered at c. The distance from the center c to the endpoints of the interval is
called the radius of convergence R. If the interval is the whole real line then R =  . If R
is finite, then the power series diverges for |x - c| > R. Convergence at the endpoints of
the interval x = c ± R must be checked separately.
Our goal is to find a power series representation for a given function f(x).
Suppose f(x) is the given function; we want to determine the {ak} so that:

f(x) =
a (xc)
k
k0
k
for |x - c| < R. By evaluating f and all its derivatives at c, it can be
f (k)(c)
shown that ak must satisfy: ak 
k!
Thus the power series for f(x) can only be the Taylor series of f(x) about x = c.
(
k
)

f
(
c
)k
2
(
x

c
)(

f
c
)

f
'
(
c
)
(
x

c
)

f
'
'
(
c
)
(
x

c
)
/
2
!

f(x) = 
k
!
k

0
When c = 0, we have a special case of Taylor series called the MacLaurin Series:
2
3
f
'
'
(0)x
f
'
'
'
(0)x
f(x)

f(0)

f
'
(0)x



.
.
.
2!
3
!
All of this assumes that the given function f(x) has a power series representation.
How can we be sure that this is the case? Here is an example of a function which is not
2
equal to its Taylor series except at x = 0: f(x) = e 1 x for x ≠ 0, and f(0) = 0. It can be
shown that all of the derivatives of f at zero are zero. Thus the Taylor series for f is the
function T(x) = 0 for all x, while f(x) is never zero when x ≠ 0.
Taylor's formula (with remainder) tells us when f(x) is equal to its Taylor series.
Suppose that f(x) has n+1 derivatives in a interval I containing the points x and c. Then
there exists a number z between x and c such that:
2
(
n
)
n
(
x
)

f
(
c
)

f
'
(
c
)
(
x

c
)

f
'
'
(
c
)
(
x

c
)
/
2
!


f
(
c
)
(
x

c
)
/
n
!

R
(
x
)
f(x) = f
n
(1
n

)
n

1
Rn ( x) is called the remainder term. Note that
(
x
)

f
(
z
)
(
x

c
)/
(
n

1
)
!
where R
n
Rn ( x) looks just like the (n+1)st term in the Taylor series except that the (n+1)st
derivative is evaluated at a point other than c. Now f(x) = Tn ( x) + Rn ( x) and f(x) is
equal to its Taylor series expansion provided that limRn(x)0.
n
