9.1 Sequences
Sequence = a function whose domain is the set of positive #’s
Ex. 1 = a1 2= a2 3= a3 4= a4 …. n = an
** a’s represent terms – can be mapped on a graph by plotting points
Recursively defined sequence – when you take the previous term and use it to find the next term
Ex. a1 = 3
ak+1 = 2(ak -1)
LIMITS OF A SEQUENCE
Definition: Let L be a real #. The limit of a sequence {an} is L written as: lim 𝑎𝑛 = 𝐿
𝑛→∞



If for each ε > 0 there exists a M>0 such that |𝑎𝑛 − 𝐿| < ε whenever n> M
If the limit L of a sequence exists then the sequence CONVERGES to L. If the limit of a sequence
does not exist then the sequence DIVERGES
Look at figure
** If a sequence {an} agrees with a function f at every positive integer, and if f(x) approaches a limit L
as x∞, the sequence must converge to the same limit L
THEOREM: LIMIT OF A SEQUENCE
a) Let L be a real #. Let f be a function of a real variable such that lim 𝑓(𝑥) = 𝐿
𝑥→∞
b) If {an} is a sequence such that f(n) = an for every positive integer n, then lim 𝑎𝑛 = 𝐿
𝑛→∞
THEOREM 5.15 A LIMIT INVOLVING e
𝟏
𝒙+𝟏 𝒙
)
𝒙→∞ 𝒙
𝐥𝐢𝐦 (𝟏 + 𝒙)𝒙 = 𝐥𝐢𝐦 (
𝒙→∞
Commonly Occurring Limits:
1.
2.
ln 𝑛
lim
𝑛→∞ 𝑛
= 0
𝑛
lim √𝑛 = 1
𝑛→∞
1
3.
4.
lim 𝑥 𝑛 = 1
𝑛→∞
(x>0)
lim 𝑥 𝑛 = 0 (|𝑥| < 1)
𝑛→∞
= e
5.
6.
𝑥
lim (1 + 𝑛)𝑛 = 𝑒 𝑥
𝑛→∞
𝑥𝑛
𝑛→∞ 𝑛!
lim
(any x)
= 0 (any x)
**Ways in which a sequence can fail to have a limit:
1) Terms of the sequence increase without bound: 𝐥𝐢𝐦 𝒂𝒏 = ∞
𝒏→∞
2) Terms of the sequence decrease without bound: 𝐥𝐢𝐦 𝒂𝒏 = −∞
𝒏→∞
Convergence Vs. Divergence
Ex. 𝑎𝑛=(−1)𝑛 (
Ex. 𝑎𝑛 =
𝑛
)
𝑛+1
3𝑛2 −𝑛+4
2𝑛2 +1
Using L’Hopital’s rule to determine Convergence
𝑛2
Ex. Show that the sequence whose nth term is an = 2𝑛 −1 converges
Factorial (n!) – defined as n! = 1·2·3·4·…·(n-1)·n
0! = 1
(2n)! vs 2n!
Squeeze Theorem for Sequences
If lim 𝑎𝑛 = L = lim 𝑏𝑛 and there exists an integer N such that an ≤ cn ≤ bn for all n>N then the
𝑛→∞
𝑛→∞
lim 𝑐𝑛 = L
𝑛→∞
Ex. 5 p. 597 in book
(−1)𝑛
𝑛!
is squeezed between 2 sequences that converge to 0
𝒌𝒏
𝒏→∞ 𝒏!
*** For any fixed #k 𝐥𝐢𝐦
= 0 *** -- the factorial function grows faster than any exponential f(x)!
Theorem 9.4 Absolute Value Theorem – if the absolute value sequence converges to 0, the original
signed sequence also converges to 0:
For sequence {an} if lim |𝑎𝑛 | = 0 then lim 𝑎𝑛 = 0
𝑛→∞
1
1
𝑛→∞
1
Ex. 1, -½, 22 , -23 , … (-1)n2𝑛 if you take the absolute value of each term, the sequence converges to 0 so:
lim [(−1)𝑛
𝑛→+∞
1
]
2𝑛
=0
Pattern Recognition for Sequences – describes the nth term after discovering a pattern
Ex. Finding the nth term for a sequence:
8 −26 80 −242
, ,
2
6 24 120
-2, ,
Steps: Look at numerator first – appears to match 3n – 1
Look at denominator next – alternates between +/- so know (-1)n
The denominator also resembles a factorial pattern so n!
Final pattern matches:
3𝑛 −1
(−1)𝑛 (𝑛!)
Now find the limit of the pattern: lim |𝑎𝑛 | = lim (−1)𝑛
𝑛→∞
𝑛→∞
3𝑛 −1
=
𝑛!
0  so converges to 0
Monotonic Sequences and Bounded Sequences

So far we have determined the convergence of a sequence by finding its limit  now we will
provide a test for convergence of sequences without determining the limit
Definition of a Monotonic Sequence – a sequence {an} is MONOTONIC if its terms are
nondecreasing (each successive term is larger) a1≤ a2≤ a3≤ …≤an or its terms are nonincreasing (each
successive term is smaller)
Look at Graphs:
**Another way to see that the sequence is monotonic is:  to argue that the derivative of the
corresponding function is positive for all x  this implies that f is increasing which implies that {an} is
increasing.
Definition of a Bounded Sequence
1. A Sequence {an} is bounded above if there is a real #M such that an ≤ M for all n. The number M
is called the UPPER BOUND of the sequence.
2. A Sequence {an} is bounded below if there is a real #N such that N ≤an for all n. The number N is
called a LOWER BOUND of the sequence.
3. A sequence {an} is bounded if it is bounded above and bounded below.
Complete Property of Real Numbers: means there are no holes or gaps on the real # line.

Used to conclude that if a sequence has an upper bound, it must have a LEAST UPPER BOUND
(an upper bound that is smaller than all other upper bounds for the sequence)
THEOREM 9.5 Bounded Monotonic Sequences

If a sequence {an} is bounded and monotonic  then it CONVERGES
Ex. an =
bn =
1
𝑛
is bounded and monotonic – so it converges
𝑛2
𝑛+1
is monotonic but not bounded (only below)
cn = (-1)n is bounded by not monotonic
Ex. Determine if the sequence with the given nth term is monotonic. Discuss the boundedness
1
#83. an= 4 - 𝑛
a1
4-1 = 3
a3 4 - ⅓ = 3.66
a2
4-½ = 3.5
a4 4 - ¼ = 3.75
** as you look at each subsequent term, it is getting larger 
SO MONOTONIC
** also see if graph – that it becomes bounded above and below
9.2 Series and Convergence
INFINITE SERIES: ∑∞
𝒏=𝟏 𝒂𝒏 = a1 + a2 + a3 + …+ an
Definitions of Convergent and Divergent Series




For the infinite series ∑∞
𝑛=1 𝑎𝑛 , the nth partial sum is given by: Sn = a1 + a2 +…+an
If the sequence of partial sums {sn} converges to S, then the series above CONVERGES
The limit S is called the sum of the series
If {sn} diverges then the series DIVERGES
Ex. Find the first 5 terms of the sequence of partial sums:
1 + ¼ + 1/9 + 1/16 + 1/25 +…+
S1 = 1
S2 = 1 + ¼ = 1.25
S3 = 1 + ¼ + 1/9 = 1.3611
S4 = 1 + ¼ + 1/9 + 1/16 = 1.4236
S5= 1 + ¼ + 1/9 + 1/16 + 1/25 = 1.4636
Ex. Convergent Series:
∑∞
𝑛=1
1
2𝑛
= ½ + ¼ + 1/8 +1/16 + …
2𝑛 −1
𝑛→∞ 2𝑛
So lim
s1 = ½ s2 = ½ + ¼ = ¾ s3 = ½ + ¼ + 1/8 = 7/8 sn=
= 1 so Converges
Ex. ∑∞
𝑛=1 1 = 1 + 1 + 1 +1
 sn = n which diverges
2𝑛 −1
2𝑛
Telescoping Series: (b1 – b2) + (b2 - b3)+ (b3 – b4) +…
**sn = b1- bn+1
** a telescoping series will converge if and only if bn approaches a finite number as n∞
** Also, if the series converges, its sum is S = b1 - lim 𝑏𝑛+1
𝑛→∞
*** If series converges then: Sum = b1 - 𝐥𝐢𝐦 𝒃𝒏+𝟏
𝒏→∞
2
Ex. Find the sum of ∑∞
𝑛=1 4𝑛2 −1
2
Step 1: Use partial fractions and rewrite as: an = 4𝑛2 −1  2 = A(2n+1) - B(2n – 1)
1
1
So an = 2𝑛−1 - 2𝑛+1
1
Sn = 2(1)−1 −
∑∞
𝑛=1
2
4𝑛2 −1
1
2(1)+ 1
+
1
1
− 2(2)+ 1 +
2(2)− 1
1
= lim ( 1 − 2𝑛+1) = 1
𝑛→∞
1
… + = 1 - 2𝑛+1
so Converges
𝒏
2
n
Geometric Series -- ∑∞
a≠0 is a geometric series with ratio r
𝒏=𝟎 𝒂𝒓 = a + ar + ar +…+ ar
Theorem: Convergence of a Geometric Series
*a geometric series with ratio r diverges is |𝑟|≥ 1. If 0 < |𝑟|< 1 then the series converges to the sum
𝒏
∑∞
𝒏=𝟎 𝒂𝒓 =
𝒂
𝟏−𝒓
0<|𝒓|<1
Ex. Verify if it diverges or converges
a)
3
n
∑∞
𝑛=0 3(2)
𝑛!
b) ∑∞
𝑛=1 2𝑛
** Make sure that you know the difference between an infinite series and a sequence:


Infinite Series = an infinite sum of terms form a sequence a1+a2+…+an
Sequence = an ordered collection of numbers a1,a2,a3…an
Theorem 9.7 Properties of Infinite Series

If ∑ 𝑎𝑛 = A and ∑ 𝑏𝑛 = B and c is areal number, then the following series CONVERGE to the
indicated series
a) ∑∞
𝑛=1 𝑐𝑎𝑛 = cA
b) ∑∞
𝑛=1(𝑎𝑛 + 𝑏𝑛 ) = A+B
∞
c) ∑𝑛=1(𝑎𝑛 − 𝑏𝑛 ) = A-B
Theorem 9.8 Limit of nth Term of a convergent series

∑∞
𝑛=1 𝑎𝑛 converges, then lim 𝑎𝑛 = 0
𝐼𝑓
**converse is generally not true
𝑛→∞
Theorem 9.9 nth term test for Divergence  states that it the limit of the nth term of a series does not
converge to 0, the series must diverge.
*If 𝐥𝐢𝐦 𝒂𝒏 ≠0 then ∑∞
𝒏=𝟏 𝒂𝒏 diverges
𝒏→∞
𝑛
Ex. ∑∞
𝑛=0 2
𝑛!
Ex. ∑∞
𝑛=1 2𝑛!+1
1
Ex. ∑∞
𝑛=1 𝑛
9.3 The Integral Test  applies only to series with positive terms
Theorem 9.10 The Integral Test

∞
If f is positive, continuous, and decreasing for x≥1 and an = f(n) then ∑∞
𝑛=1 𝑎𝑛 and ∫1 𝑓(𝑥)𝑑𝑥
either both converge or both diverge
2
Ex. ∑∞
𝑛=1 3𝑛+5
2
Let f(x) = 3𝑛+5
x=1
2
3(1)+ 5
2
=8
x=2
2
3((2)+ 5
 +/Contin. And decreasing
∞
Integrate so: ∫1
#8.
ln 2
√2
+
ln 3
√3
∑∞
𝑛=2
ln 𝑛
√𝑛
+
ln 4
√4
+
ln 5
√5
2
3𝑥+5
2
𝑑𝑥 = 3 [ln(3𝑥 + 5)]∞1 = ∞ Diverges
+…
ln 𝑥
√𝑥
so f(x) =
Step #1. Derive f´(x)=
2−ln 𝑥
3
 f is positive and continuous, and decreasing for x>2
2𝑥 2
∞ ln 𝑛
Step #2. Integrate ∫2
#11 ∑∞
𝑛=1
√𝑥
dx = [2√𝑥 (ln 𝑥 − 2)]∞2 = ∞ DIVERGES
1
√𝑛+1
P-Series and Harmonic Series - has a simple arithmetic test for convergence or divergence
𝟏
P-Series ∑∞
𝒏=𝟏 𝒏𝒑 =
𝟏
𝟏𝒑
𝟏
𝟏
+ 𝟐𝒑+ 𝟑𝒑 + … is a p-series where p is a positive constant
1
For p = 1 ∑∞
𝑛=1 𝑛 = 1+ ½ + 1/3 +…+ = HARMONIC SERIES

1
General Harmonic Series is of the form ∑ (𝑎𝑛+𝑏)
1
1
1
1
Theorem 9.11 Convergence of P-Series: The p-series ∑∞
𝑛=1 𝑛𝑝 = 1𝑝 + 2𝑝 + 3𝑝 + …


Converges if p>1
Diverges if 0 < p ≤1
Ex. discuss the convergence/divergence of:
a) Harmonic series ∑∞
𝑛=1
b) P series with p = 2
1
𝑛
= 1 + ½ + 1/3 +…
1
∑∞
𝑛=1 12
+
1
1
+
22 32
+…
Testing for Convergence
∑∞
𝑛=2
1
𝑛 ln 𝑛

1
𝑥 ln 𝑥
is positive and continuous for x≥2
DIVERGES b/c ≤ 1
p>1 so CONVERGES
1+ln 𝑥
Steps: 1. find derivative (𝑥 ln 𝑥)−1  f´(x) = (-1)(𝑥 ln 𝑥) −2(1+ln x) = − 𝑥 2 (ln 𝑥 2 )
2. f´(x)<0 for x>2 so now do integral test
∞ 1
∫2 𝑥 ln 𝑥
1
𝑑𝑥 =
∞
∫2 ln𝑥𝑥
dx = lim [ln(ln 𝑥)]b2  lim [ln(𝑙𝑛𝑏) − ln(𝑙𝑛2] = ∞
𝑏→∞
𝑏→∞
9.4 Comparisons of Series  positive term series

Past convergence tests
a) Terms of series have to be fairly simple
b) Series must have special characteristics in order for convergence tests to be applied
 A slight deviation from these special characteristics can make a test nonapplicable
**the next tests allow you to compare a series having complicated terms with a simple series whose
convergence or divergence is known.
Theorem 9.12 Direct Comparison Test
Let 0 < an ≤bn for all n
∞
1. If ∑∞
𝒏=𝟏 𝒃𝒏 converges, then ∑𝒏=𝟏 𝒂𝒏 converges
∞
2. If ∑∞
𝒏=𝟏 𝒂𝒏 diverges, then ∑𝒏=𝟏 𝒃𝒏 diverges
1
Ex. ∑∞
𝑛=1 3𝑛2 +2
Ex. ∑∞
𝑛=1
1
3
4 √𝑛−1
1
1
(this is similar to 3𝑛2 )  like 1/3∑∞
𝑛=1 𝑛2 which is a p series where p>1 so Converge
1
compare with p series: 1/4∑∞
𝑛=1 4
√𝑛
diverges b/c p<1 (1/4)
Limit Comparison Test – a given series closely resembles a p-series or a geometric series, but you cannot
establish the term by term comparison necessary to apply the Direct Comparison Test  so you use
now a LIMIT COMPARISON TEST
Theorem 9.13 LIMIT COMPARISON TEST
𝑎
*Suppose that an>0, bn>0 and lim (𝑏𝑛 ) = L
𝑛→∞
𝑛
L is finite and positive. Then the two series ∑ 𝑎𝑛
and ∑ 𝑏𝑛 either both converge or both diverge.
1
Ex. Show that ∑∞
𝑛=1 𝑎𝑛+𝑏 a>0 and b>0

By direct comparison ∑∞
𝑛=1
1
𝑛
(a divergent harmonic series)

You have lim
𝑛→∞
1
𝑎𝑛+𝑏
1
𝑛
𝑛
𝑎𝑛+𝑏
𝑛→∞
= lim
1
=𝑎
 because the limit is greater than 0, you can conclude
the given series diverges
A LIMIT COMPARISON TEST WORKS WELL FOR COMPARING A “MESSY” ALGEBRAIC SERIES WITH A PSERIES!
*In choosing an appropriate p-series, you must choose one with an nth term of the same magnitude as
the nth term of the given series.
Given Series
Comparison Series
Conclusion
∑∞
𝑛=1
1
3𝑛2 −4𝑛+5
∑∞
𝑛=1
1
𝑛2
Both Series Converge
∑∞
𝑛=1
1
√3𝑛−2
∑∞
𝑛=1
1
√𝑛
Both Series Diverge
∑∞
𝑛=1
𝑛2 −10
4𝑛5 + 𝑛3
∑∞
𝑛=1
𝑛2
𝑛5
1
= ∑∞
𝑛=1 𝑛3
Both series converge
*** When choosing a series for comparison, you can disregard all but the highest powers of n in both
the numerator and denominator ***
Ex. Determine the convergence or divergence of:
√𝑛
𝑛2 + 1
a)
∑∞
𝑛=1
b)
∑∞
𝑛=1
c)
∑∞
𝑛=1
𝑛2𝑛
4𝑛3 +1
1
𝑛√𝑛2 +1
9.5 Alternating Series


Series that contain both positive and negative terms
Alternating series occur in 2 ways: the odd terms are negative or the even terms are negative
Theorem 9.14 Alternating Series Test:
Let an>0. The Alternating Series
following 2 conditions are met:
1)
lim 𝑎𝑛 = 0
𝑛→∞
and
∞
𝑛
𝑛+1
∑∞
𝑎𝑛 Converge if the
𝑛=1(−1) 𝑎𝑛 and ∑𝑛=1(−1)
2) 𝑎𝑛+1 ≤ an for all n
1
𝑛+1
Ex. Determine the convergence or divergence of ∑∞
𝑛=1(−1)
𝑛
a)
lim 𝑎𝑛 
𝑛→∞
b) An+1 =
1
𝑛+1
1
𝑛
𝑛→∞
1
𝑛
lim
≤
= converges to 0
Ex. determine the convergence/divergence of
∞
∑
𝑛=1
∞
∑
𝑛=1
(−1)𝑛+1
2𝑛 − 1
(−1)𝑛+1 ln(𝑛 + 1)
𝑛+1
p. 632 are examples where alternating series test fails
ALTERNATING SERIES REMAINDER – the partial sum Sn can be a useful approximation for sum S of the
series. The error involved in using S≈Sn is the remainder Rn = S - Sn

If a convergent alternating series satisfies the condition an+1≤an, then the absolute value of the
remainder Rn involved in approximating the sum S by Sn is less than or equal to the first
|𝑆 − 𝑆𝑛 | = |𝑅𝑛 | ≤ an+1
neglected term
Ex. ∑∞
𝑛=1
(−1)𝑛+1 3
𝑛2
approximate the sum of the series by using the 1st 6 terms
Absolute and Conditional Convergence
**Sometimes a series may have both + and – terms and not be an alternating series  so to obtain
information about the convergence – you must investigate (by direct comparison, etc…)
Theorem 9.16 Absolute Convergene


If the series ∑|𝑎𝑛 | converges, then the series ∑ 𝑎𝑛 also converges.
Converse of theorem 9.16 is not true
Ex. Alternating harmonic series ∑∞
𝑛=1
(−1)𝑛+1
𝑛
1
2
1
3
1
4
= 1- + - +…
Definition of Absolute and Conditional Convergence
1. ∑ 𝒂𝒏 is absolutely convergent if ∑|𝒂𝒏 | converges
2. ∑ 𝒂𝒏 is conditionally convergent if ∑ 𝒂𝒏 converges but ∑|𝒂𝒏 | diverges
Ex. Determine whether the series converges conditionally or absolutely diverges
#49. ∑∞
𝑛=1
So
(−1)𝑛+1
√𝑛
=
−1
√𝑛+1
1
√𝑛
<
However ∑∞
𝑛=1
(−1)1+1
√1
+
(−1)2+1
…
√2
1
𝑛→∞ √𝑛
and lim
1
√𝑛
=0
1
1
√2
=1 -
so by alternating series test CONVERGES
(p=1/2 which is p<1  so diverges)
***SO SERIES CONVERGES CONDITIONALLY ***
#58. ∑∞
𝑛=0
(−1)𝑛
√𝑛+4
Rearrangement of Series  in infinite series the value of the sum may change depending on whether
the series is absolutely convergent or conditionally convergent.
1
1
𝑛+1
Ex. the alternating harmonic series converges to ln 2 ∑∞
=1–½+3-4+…
𝑛=1(−1)
**however, rearrangement of the series produces a different sum:
9.6 The Ratio and Root Tests
Test for absolute convergence:
Let ∑ 𝑎𝑛 be a series with nonzero terms
RATIO TEST:
1.
2.
3.
𝑎𝑛+1
|<1
𝑎𝑛
𝑎
𝑎
∑ 𝑎𝑛 diverges if lim | 𝑛+1 | >1 or lim | 𝑛+1 | = ∞
𝑎
𝑛→∞
𝑛→∞ 𝑎𝑛
𝑛
𝑎
The ratio test is inconclusive if lim | 𝑎𝑛+1 | = 1
𝑛→∞
𝑛
∑ 𝑎𝑛 converges ABSOLUTELY if lim |
𝑛→∞
**Ratio Test  is particularly useful for series that converge rapidly (series involving
factorials/expoentials)
Ex. ∑∞
𝑛=0
2𝑛
𝑛!
𝑛!
𝑛!
1
=
=
(𝑛+1)! (𝑛+1)𝑛! 𝑛+1
2𝑛+1
2𝑛
lim [(𝑛+1)! ÷ 𝑛! ]
𝑛→∞

Simplifying quotients of factorials:

An =
2𝑛
𝑛!
write as lim |
𝑛→∞
𝑎𝑛+1
|
𝑎𝑛
=
2𝑛+1
𝑛!
= lim [(𝑛+1)! ∙ 2𝑛 ]
𝑛→∞
= lim
2
𝑛→∞ 𝑛+1
Ex. a) ∑∞
𝑛=0
𝑛2 2𝑛+1
3𝑛
= 0 Coverges
𝑛𝑛
c) ∑∞
𝑛=1 𝑛!
THE ROOT TEST – works well for series involving nth powers
Theorem 9.18 Root test:
Let ∑ 𝒂𝒏 be a series:
𝒏
1. ∑ 𝒂𝒏 converges absolutely if 𝐥𝐢𝐦 √|𝒂𝒏 | <1
𝒏→∞
𝒏
𝒏
2. ∑ 𝒂𝒏 diverges if 𝐥𝐢𝐦 √|𝒂𝒏 | >1 or 𝐥𝐢𝐦 √|𝒂𝒏 | = ∞
𝒏→∞
𝒏→∞
𝒏
3. The root test is inconclusive if 𝐥𝐢𝐦 √|𝒂𝒏 | = 1
𝒏→∞
**The root test is always inconclusive for any p-series **
Ex. Determine the convergence or divergence of: ∑∞
𝑛=1
𝑛
𝑒 2𝑛
𝑛𝑛
𝑒 2𝑛
𝑛
lim √|𝑎𝑛 | = lim √ 𝑛𝑛
𝑛→∞
𝑛→∞
2𝑛
= lim
𝑛→∞
𝑒𝑛
𝑛
𝑛𝑛
𝑒2
𝑛→∞ 𝑛
= lim
= 0 which is < 1  this limit will always be <1 so it
Converges absolutely.
Guidelines for Testing a Series for Convergence or Divergence:
1. Does the nth term approach 0? If not  the series DIVERGES
2. Is the series one of the special types  geometric, p-series, telescoping, or alternating
3. Can the Integral, Root or Ratio Test be applied
4. Can the series be compared favorably to one of the special types.
Look at example #5 p. 643 and make Chart!
9.7 Taylor Polynomials and Approximations p(x) = a0 + a1x + a2x2 + a3x3 + … + anxn
*local linear approximation of 1st degree polynomial: f(x) ≈ f(x0) + f´(x0)(x – x0)
Polynomial Approximations of elementary functions
Goal: To use polynomial functions as approximations of other functions
Must: #1. 1st begin by choosing a #c in the domain of f at which f and P have the same value where
P(c ) = f(c)  the graphs of both will pass through (c, f(c))
** approximating polynomial is said to be expanded about c or centered at c
** Many polynomial graphs pass through the point (c, f(c))
** YOUR TASK: to find the polynomial whose graph resembles the graph of f at this pt.
#2. To do this: ignore the additional requirement that the slope of the polynomial function be
The same as the slope of the graph of f at the point (c,f(c))  P´(c) = f´(c)
Ex. For function f(x) = ex find the first degree polynomial function P1(x) = a0 + a1x whose value and slope
agree with the value and slope of f at x=0.




Because f(x) = ex
f´(x) = ex
f(0) = e0 = 1
f´(0) = e0 = 1  The value of f and the slope of f at x = 0
Because P1(x) = a0 + a1x you can use the condition P1(0) = f(0) so a0 = 1
Also because P1´(x) = a1 you can use the condition that P1´(0) = f´(0) so P1´ = 1
So = P1(x) = 1 + x  at (0,1) the graphs are reasonably close – but as you move away the
graphs move farther away from each other  so need to improve approximation even more by:
** that the values of the 2nd derivatives of P and f agree when x=0 has the form
P(x) ≈ a0 + a1x + a2x2
P(0) = f(0) = 1
p´(0) = f´(0) = 1x
p´´(0) = f´´(0) = ½ x²  p´´(x) = 2a2 so p´´(0) = 2(0) = 0  p2(x) = 1 + x + ½ x2
**if you continued finding the derivatives you would get:
𝟏
𝟑!
𝟏
𝒏!
Pn(x) = 1 + x + ½ x2 + x3 + …+ xn ≈ ex
Taylor and Maclaurin Polynomials
**for expansion about an arbitrary value of c write the polynomial in the form:
Pn(x) = a0 + a1(x-c) + a2(x-c)2 + a3(x-c)3 + …+ an(x-c)n
Pn´(x) = a1 + 2a2(x-c) + 3a3(x-c)2 + … + nan(x-c)n-1
Pn´´(x) = 2a2 + 2(3a3)(x-c) + … + n(n-1)an(x-c)n-2
Pn´´´(x)= 2(3a3) + …+ n(n-1)(n-2)an(x-c)n-3 . . .
Pnn(x) = n(n-1)(n-2) . . . 2(1)an
So letting x=c you get… Pn(c) = a0 Pn´(c) = a1 Pn´´9c) = 2a2 pnn(c) = n!an
**Since the value of f and its first n derivatives must agree with the value of P n and its first n derivatives
at x=c follows that: f(c) = a0 f´(c) = a1
𝑓´´ (𝑐)
2!
= a2
𝑓𝑛 (𝑐)
𝑛!
= 𝑎𝑛
Taylor Polynomial for f at c – if f has n derivatives at c, then the polynomial:
Pn(x) = f(c) + f´(c)(x-c) +
𝑓´´ (𝑐)
2!
(x-c)2 +. . . +
𝑓(𝑛) 𝑐
𝑛!
(x-c)n
Maclaurin Polynomial for f if c = 0
Pn(x) = f(0) + f´(0)x +
𝑓´´ (0) 2
x
2!
+
𝑓´´´ (0) 3
x
3!
+...+
𝑓(𝑛) (0) n
x
𝑛!
Ex. Find the Taylor polynomials P0, P1, P2, P3, P4 for f(x) =lnx centered at c=1
F(x) = ln x
f(1) = ln 1 = 0
1
1
f´(x) = 𝑥
f´(1) = 1 = 1
1
f´´(x) = − 𝑥 2
2!
f´´´(x) = 𝑥 3
3!
f(4)(x) = -𝑥 4
f´´(1) =
−1
12
= -1
2!
f´´´(1) = 13 = 2
f(4)(1) =
−3!
=
14
-6
Taylor Polynomials are as follows:
P0(x) = f(1) = 0
P1(x) = f(1) + f´(1)(x-1) = (x-1)
P2(x) = f(1) + f´(1)(x-1) +
𝑓´´ (1)
(x-1)2
2!
= (x-1) – ½ (x-1)2
P3(x) = f(1) + f´(1)(x-1) +
𝑓´´ (1)
(x-1)2
2!
+
𝑓´´´ (1)
(x-1)3
3!
= (x-1) – ½ (x-1)2 + 1/3(x-1)3
P4(x) = f(1) + f´(1)(x-1) +
𝑓´´ (1)
(x-1)2
2!
+
𝑓´´´ (1)
(x-1)3
3!
+
𝑓4 (1)
(x-1)4
4!
= (x-1) – ½ (x-1)2 + 1/3(x-1)3 – ¼(x-1)4
Find the Maclaurin polynomial P0, P2, P4, and P6 for f(x) = cos x use P6(x) to approximate the value of
cos(0.1)
F(x) = cos x
f(0) = cos 0 =1
f´(x) = - sinx
f´(0) = -sin0 = 0
f´´(x) = -cosx
f´´(0)= -cos0 = -1
f´´´(x) = sinx
f´´´(0) = sin0 = 0
P0(x) = 1
P2(x) = 1 + 0 +
𝑓´´ (0) 2
x
2!
 1 – ½!x2
P4(x) = 1 + 0 +
𝑓´´ (0) 2
x
2!
+
𝑓´´´ (0) 3
x
3!
+
𝑓4 (0) 4
x
4!
 1 – ½!x2 + ¼!x4
P6(x) 1 – ½!x2 + ¼!x4 – 1/6!x6
**Using P6 to approximate cos(.1) = 1- ½!(.1)2 + ¼!(.1)4 – 1/6!(.1)6 = .995004165
Look at p. 652 Taylor Poly. For sin(x) expanded about π/6
**Taylor and Maclaurin Polynomials can be used to approximate the value of a function at a specific
point.
Ex. to approximate the value of ln(1.1)  use taylor poly for f(x) = ln x expanded about c=1
Ex. Use a fourth maclaurin poly to approximate the value of ln(1.1)  (1.1 is closer to 1 than 0 you use
Mauclaurin poly for function g(x) = ln (1+x)
g(x) = ln(1+x)
g´(x) =
g(0) = ln(1+0) = ln 1 = 0
1
(1+x)-1
1+𝑋
g´(0) = (1+0)-1 = 1
g´´(x) = -(1+x)-2
g´´(0) = -(1+0)-2 = -1
g´´´(x) = 2(1+x)-3
g´´´(0) = 2(1+0)-3 = 2
g4(x) = -6(1+x)-4
g4(0) = -6(1+0)-4 = 6
P4(x) = g(0) + g´(0) +
𝑔´´ (0) 2
x
2!
+
𝑔´´´ (0) 3
x
3!
+
𝑔4 (0) 4
x
4!
= x – ½ x2 + 1/3x3 – ¼ x4
so ln(1.1) = ln(1+.1)≈P4(.1) = .0953083
Remainder of a Taylor Polynomial
**To measure the accuracy of approximating a function value f(x) by the Taylor Polynomial P n(x) you can
use the concept of a remainder Rn(x):
f(x) = Pn(x) + Rn(x) remainder
exact value
approximate value
so Rn(x) = f(x) – Pn(x)  the absolute value of Rn(x) is called ERROR associated with approximation
ERROR = |𝑅𝑛 (𝑥)| = |𝑓(𝑥) − 𝑃𝑛 (𝑥)|
TAYLOR THEOREM: if a function f is differentiable through order n+1 in an interval I containing c, then
for each x in I there exists Z between x and c such that:
F(x) = f(c)+ f´(c)(x-c) +
𝑓´´ (𝑐)
(x-c)2 +
2!
...+
𝑓(𝑛) (𝑐)
(x-c)n
𝑛!
+ Rn(x) where Rnx =
𝑓(𝑛+1) (𝑧)
(x-c)n+1
(𝑛+1)!
**one useful consequence of Taylor’s Theorem is that
|𝑅𝑛 (𝑥)|≤
|𝑥−𝑐|𝑛+1
(𝑛+1)!
max |𝑓 𝑛+1 (𝑧)| where max |𝑓 𝑛+1 (𝑧)| is the maximum value of f(n+1)(z) between x & c
Ex. The 3rd Maclaurin polynomial for sinx is given by P3(x) = x -
𝑥3
.
3!
Use the Taylor Polynomial Theorem
to approximate sin(.1) by P3(.1) and determine the accuracy of the approximation.
𝑥3
Sin x = x - 3! + R3(x) = x So sin(.1) ≈ .1 –
(.1)3
3!
𝑥3
3!
+
𝑓4 (𝑧) 4
x
4!
where 0 < z < .1
 .1 - .000167 = .099833
Because f4(z) = sin z it follows that the error |𝑅3 (.1)| can be bounded as follows:
0<R3(.1) =
sin 𝑧
(.1)4
4!
<
.0001
4!
≈ .000004
So this implies: .099833 < sin (.1) = .099833 + R3(x) < .099833 + .000004  .099833 <sin.1 < .099837
Ex. Determine the degree of the Taylor Polynomial Pn(x) expanded about c=1 that should be used to
approximate ln (1.2) so that the error is less than .001
𝑛!
F(x) = lnx and the (n+1)st derivative is givne by f(n+1)(x) = (-1)n𝑥 𝑛+1
𝑓𝑛+1 (𝑧)
**Using Taylor’s Theorem |𝑅𝑛 (1.2)| = | (𝑛+1)! (1.2 − 1)𝑛+1 | =
=
** In this interval,
.2𝑛+1
[𝑧 𝑛+1 (𝑛+1)]
is <
.2𝑛+1
𝑍𝑛+1 (𝑛+1)
𝑛!
1
[
]
𝑧 𝑛+1 (𝑛+1)!
.2n+1
where 1 < z < 1.2
.2𝑛+1
(𝑛+1)
.2𝑛+1
(𝑛+1)
So you are seeking a value of n such that:
< .001  1000 < (n+1)5n+1
So the smallest value n that satisfies the inequality is n=3 so you would need the 3 rd Taylor Poly. To
achieve desired accuracy.
9.8 Power Series
Approximating Functions by Taylor/Maclaurin Polynomials
Ex. f(x) = ex  1st degree ex≈ 1+x
2nd degree ex≈ 1+ x +
𝑥2
2!
3rd degree ex≈ 1 + x +
𝑥2
2!
+
𝑥3
3!
4th degree ex ≈ 1 + x+
𝑥2
2!
+
𝑥3
3!
+
𝑥2
2!
+
𝑥3
3!
+
5th degree ex ≈ 1 + x +
𝑥4
4!
𝑥4
4!
+
𝑥5
5!
**the higher degree of the approximating polynomial, the better the approximation becomes
**Several important functions, such as f(x) = ex can be represented exactly by an infinite series called 
POWER SERIES
 In example above, f(x) = ex  for each real # x, it can be shown that the infinite series:
ex = 1 + x +
𝑥2
2!
+
𝑥3
3!
+…+
𝑥𝑛
𝑛!
converges on the right to ex
Definition of Power Series – If x is a variable, then an infinite series of the form:
𝒏
2
3
n
∑∞
𝒏=𝟎 𝒂𝒏 𝒙 = a0 + a1x + a2x + a3x + . . . + anx + … is called a POWER SERIES.
𝒏
2
n
More generally, an infinite series of the form: ∑∞
𝒏=𝟎 𝒂𝒏 (𝒙 − 𝒄) = a0 + a1(x-c) + a2(x-c) +…+an(x-c) + …
is called a POWER SERIES CENTERED AT C, where c is a constant.
Ex. p. 660
Radius and Interval of Convergence:
𝒏
*A power series in x can be viewed as a function of x  f(x) = ∑∞
𝒏=𝟎 𝒂𝒏 (𝒙 − 𝒄)
DOMAIN of f: the set of all x for which the power series converges

Every power series converges at its center  so c always lies in the domain of f
The Domain of a power series can take 3 forms:
1. A single point
2. An interval centered at c
3. An entire real line
Theorem 9.20 Convergence of a Power Series:
For a power series centered at c, precisely one of the following is true:
1) The series converges only at c
2) There exists a real # R>0 such that the series converges absolutely for |(𝑥 − 𝑐)| < R and
diverges for |(𝑥 − 𝑐)| > R
3) The series converges absolutely for all x
*The # R = Radius of Convergence of the power series.
a) If the series converges only at c, the radius of convergence is R = 0
b) If the series converges for all x, the radius of convergence is R = ∞
**Interval of Convergence = the set of all values of x for which the power series converges
𝑛
n
Ex. Finding Radius of Convergence of ∑∞
𝑛=0 3(𝑥 − 2) for x≠ 2, let un = 3(x-2) then:
𝑢𝑛+1
|
𝑛→∞ 𝑢𝑛
lim |
3(𝑥−2)𝑛+1
|
𝑛→∞ 3(𝑥−2)𝑛
= lim |
= lim |𝑥 − 2| = |𝑥 − 2| by ratio test, the series converges if |𝑥 − 2|
𝑛→∞
< 1 and diverges if |𝑥 − 2| > 1. Therefore Radius of Convergence with series is R =1
Ex.
(−1)𝑛 𝑥 2𝑛+1
∑∞
𝑛=0 (2𝑛+1)!
𝑥2
(2𝑛+3)(2𝑛+2)
𝑛→∞
= lim
Let
(−1)𝑛 𝑥 2𝑛+1
un= (2𝑛+1)!
𝑢
then lim | 𝑛+1 |
𝑛→∞ 𝑢𝑛
= lim |
𝑛→∞
(−1)𝑛+1 𝑥2𝑛+3
(2𝑛+3)!
(−1)𝑛 𝑥2𝑛+1
(2𝑛+1)!
|
* for any fixed value of x, this limit is 0. So by the Ratio Test, the series converges
for all x, so Radius of Convergence = ∞
ENDPOINT CONVERGENCE:

1.
2.
3.
4.
5.
6.
Each endpoint must be tested separately for convergence or divergence  as a result, the
interval of convergence of a power series can take any 1 of the 6 forms below:
Radius: 0
Radius: ∞
Radius: R
Radius: R
Radius: R
Radius: R
Finding the Interval of Convergence:
𝑥𝑛
∑∞
𝑛=1 𝑛
letting un =
𝑥𝑛
𝑛
𝑢
 lim | 𝑛+1 |
𝑛→∞ 𝑢𝑛
= lim |
𝑛→∞
𝑥𝑛
(𝑛+1)
𝑥𝑛
𝑛
𝑛𝑥
| = lim |𝑛+1| = |𝑥|<1
𝑛→∞
** so by ratio test, the radius of convergence is R= 1
1) Because the series is centered at 0, it converges in the interval (-1,1)
**however, not necessarily the interval of convergence  to determine, you must test for the
convergence at each endpoint
1
-
When x = 1 you obtain the divergent harmonic series ∑∞
𝑛=1 𝑛
-
When x = -1 you obtain the convergent alternating harmonic series ∑∞
𝑛=1
(−1)𝑛
𝑛
-
So the interval of convergence for the series is [-1,1)
Ex. Find the interval of convergence of ∑∞
𝑛=1
𝑢𝑛+1
|
𝑛→∞ 𝑢𝑛
lim |
= lim |
𝑛→∞
(−1)𝑛+1 (𝑥+1)𝑛+1
2𝑛+1
(−1)𝑛 (𝑥+1)𝑛
2𝑛
(−1)𝑛 (𝑥+1)𝑛
2𝑛
2𝑛 (𝑥+1)
|
𝑛→∞ 2𝑛+1
| = lim |
=|
𝑥+1
|
2
un =
(−1)𝑛 (𝑥+1)𝑛
2𝑛
𝑥+1
| <1
2

By ratio test the sequence converges if |

R=2
Because the series is centered at x=-1, it will converge in the interval (-3,1) so at the endpoints
< 1 or |𝑥 + 1| < 2, so the radius of convergence is
(−1)𝑛 −2𝑛
2𝑛
∑∞
=
= ∑∞
𝑛=0
𝑛=0 1 diverges when
𝑛
2
2𝑛
𝑛 2𝑛
(−1)
𝑛
∑∞
= ∑∞
𝑛=0
𝑛=0(−1) diverges when x = 1
2𝑛
you have: ∑∞
𝑛=0

And

Both of which diverge, so the interval of convergence (-3,1)
x = -3
** HINTS: a) if all n’s cancel out, then x will have a radius
b)If n’s don’t cancel – higher degree on top then ∞  then R = 0
c) if n’s don’t cancel – higher degree on bottom then 0  then R = ∞
Differentiation and Integration of Power Series
Theorem 9.21 Properties of Functions defined by Power Series:
𝑛
If the function given by f(x) = ∑∞
= a0 + a1(x-c) + a2(x-c)2 + … has a radius of
𝑛=0 𝑎𝑛 (𝑥 − 𝑐)
convergence of R > 0, then on the interval (c-R,c+R), f is differentiable (and therefore continuous).
Moreover, the derivative and antiderivative of f are as follows:
𝑛−1
1. f´(x) = ∑∞
= a1 + 2a2(x-c) = 3a3(x-c)2 + …
𝑛=1 𝑛𝑎𝑛 (𝑥 − 𝑐)
2. ∫ 𝑓(𝑥)𝑑𝑥 = C + ∑∞
𝑛=0 𝑎𝑛
(𝑥−𝑐)𝑛+1
𝑛+1
(𝑥−𝑐)2
(𝑥−𝑐)3
+
a
+
2
2
3
= C + a0(x-c) + a1
…
Radius of Convergence of Series  obtained by differentiating or integrating is the same as original
power series
Interval of Convergence  may differ as a result of behavior of endpoints.
Ex. Consider the function given by f(x) = ∑∞
𝑛=1
𝑥𝑛
𝑛
=x+
𝑥2
2
+
𝑥3
3
Find the intervals of convergence for each of the following:
a)
∫ 𝑓(𝑥)𝑑𝑥
b) f(x)
c) f´(x)
+ ….
𝑥 𝑛+1
𝑥2
𝑥3
𝑥4
And ∫ 𝑓(𝑥)𝑑𝑥 = C + ∑∞
𝑛=1 𝑛(𝑛+1) = C + 1∙2 + 2∙3 + 3∙4 + …
By the ratio test, you can show that each series has a radius of convergence of R=1. Considering the
interval (-1,1) you have the following:
𝑥 𝑛+1
a) For ∫ 𝑓(𝑥)𝑑𝑥 the series ∑∞
𝑛=1 𝑛(𝑛+1) converges for all x = +/-1 and its I of C is [-1,1]
𝑥𝑛
converges for x =
𝑛
∞
𝑛−1
∑𝑛=1 𝑥
diverges for x
b) For f(x) the series ∑∞
𝑛=1
-1 and diverges for x=1 so I of C is [-1,1)
c) For f´(x) the series
= +/-1 and its I of C is (-1,1)
**f´(x) is the least likely to converge at the endpoints
 It can be shown that if the series for f´(x) converges t the endpoints x = c+/- R, the series
for f(x) will also converge there
9.9 Geometric Power Series
𝑛
Remember: Sum of Geometric Series ∑∞
𝑛=0 𝑎𝑟 =
𝑎
1−𝑟
|𝑟| < 1
**In other words if you let a = 1 and r = x, a power series representation for (
1
1−𝑥
1
)
1−𝑥
centered at 0 is:
𝑛
2
3
= ∑∞
𝑛=0 𝑥 = 1 + x + x + … |𝑥| < 1
1
This series represents f(x) = (1−𝑥) only on the interval (-1,1) whereas f is defined for all x≠1
**To represent f in another interval, you must develop a different series 
1
1
1
Ex. to get power series centered at -1 --> 1−𝑥 = (1−𝑥)−1 = 2−(𝑥+1)=
Which implies a= ½ r =
1
1−𝑥
1 𝑥+1 𝑛
)
2
= ∑∞
𝑛=0 2 (
𝑥+1
2
 ½ [1 +
1
2
𝑥+1)
1−[
]
2
so for |𝑥 + 1| <2 you have:
(𝑥+1)
2
+
(𝑥+1)2
4
+
(𝑥+1)3
8
+ …] |𝑥 + 1| < 2 converges on interval (-3,1)
4
Ex. Find a power series for f(x) = 𝑥+2 centered at 0
𝑎
a) Write in 1−𝑟 format =
4
4
2
2 𝑥
2+ 2
=
𝑎
= 1−𝑟
2
1−
−𝑥
2
𝑎
= 1−𝑟
a = 2 r=
∞
𝑛
b) So power series is 𝑥+2 = ∑∞
𝑛=0 𝑎𝑟 = ∑𝑛=0 2
−𝑥𝑛
2
−𝑥
2
𝑥
= 2( 1 - 2 +
𝑥2
4
-
𝑥3
+
8
…)
−𝑥
Series converges when | 2 | < 1 and implies interval of convergence = (-2,2)
Operations with Power Series: let f(x) = ∑ 𝑎𝑛 𝑥 𝑛 and g(x) = ∑ 𝑏𝑛 𝑥 𝑛
𝑛 𝑛
1. f(kx) = ∑∞
𝑛=0 𝑎𝑛 𝑘 𝑥
𝑛𝑁
2. f(xN) = ∑∞
𝑛=0 𝑎𝑛 𝑥
∞
3. f(x) + g(x) = ∑𝑛=0(𝑎𝑛 ± 𝑏𝑛 )𝑥 𝑛
** All of these operators can change the interval of convergence for the resulting series
Ex. for addition  the interval of convergence as the Intersection of the IOC’s of the original series
𝑥
∞
∞
𝑛
𝑛
∑∞
𝑛=0 𝑥 + ∑𝑛=0(2 ) = ∑𝑛=0(1 +
1
)𝑥 𝑛
2𝑛
(-1,1) Ω (-2,2) = (-1,1)
Ex. Adding 2 Power Series:
Find a power series centered at 0 for f(x) =
3𝑥−1
𝑥 2 −1
3𝑥−1
𝑥2− 1
Step 1: Use partial fractions to break down:
Step 2: Find each power series
1
𝑥−1
So
3𝑥−1
𝑥 2 −1
2
𝑥+1
1
2
1−(−𝑥)
𝑛 |𝑥|
= 1−𝑥 = -∑∞
<1
𝑛=0 𝑥
𝑛
𝑛
= ∑∞
𝑛=0[2(−1) − 1]𝑥
|𝑥| < 1
=
and x > -1
2
1
= (𝑥+1) + (𝑥−1)
𝑛 𝑛
= ∑∞
𝑛=0 2(−1) 𝑥
|𝑥| < 1
a = 1 r = -x
= 1 – 3x + x2 – 3x3 + x4… IOC is (-1,1)
(-1,1)
Ex..Finding a power series by integration
--Find a power series for f(x) = ln x centered at 1
1
𝑥
𝑛
𝑛
= ∑∞
-- a = 1 r = (1-x) -(x-1)
𝑛=0(−1) (𝑥 − 1)
1
[1−(−𝑥+1)]
1
𝑛
--Integrating produces lnx = ∫ 𝑥 𝑑𝑥 + 𝐶 = C + ∑∞
𝑛=0(−1)
(𝑥+1)𝑛+1
𝑛
-- letting x =1, then C = 0 so: ln = ∑∞
𝑛=0(−1) (
-- Interval of Convergence = (0,2]
|−(𝑥 − 1)|< 1
-x + 1 > -1
𝑛+1
) =
𝑥−1
1
(𝑥−1)𝑛+1
𝑛+1
–
(𝑥−1)2
2
+
(𝑥−1)3
3
–
(𝑥−1)4
4
+…
x>0
x<2
***Series actually converges at 2
Ex. Find a power series for g(x) = arctanx centered at 0
𝑑𝑦
[arctan 𝑥]
𝑑𝑥
1.
1
= (1+ 𝑥2 ) so use the following:
1
𝑛 𝑛
F(x) = 1+𝑥 ∑∞
𝑛=0(−1) 𝑥
IOC (-1,1)
a = 1 r = -x
1
2𝑛
2. Substitute x2 for x to get: f(x2) = 1+𝑥 2 = ∑∞
- then integrate
𝑛=0(−1)𝑥
Get: arctanx = ∫
1
1+𝑥 2
dx + C
𝑛
= C + ∑∞
𝑛=0(−1)
𝑛
= ∑∞
𝑛=0(−1)
|−𝑥| < 1
𝑥 2𝑛+1
2𝑛+1
𝑥 2𝑛+1
2𝑛+1
-x > -1
when x =0 C=0
= x-
𝑥3
3
+
𝑥5
5
-
𝑥7
7
IOC = (-1,1)
(-1,1) actually converges for x±1
9.10 Taylor Series and MacLaurin Series
Theorem 9.22 The Form of a Convergent Power Series
** If f is represented by a power series f(x) = ∑ 𝑎𝑛 (𝑥 − 𝑐)𝑛 for all x in an open interval I containing c,
then an =
𝑓(𝑛) (𝑐)
𝑛!
and f(x) = f(c) + f’(c)(x-c) +
𝑓′′ (𝑐)
(x-c)2+
2!
…+
𝑓(𝑛) (𝑐)
𝑛!
(x-c)n + …
**coefficients are the same as Taylor Polynomial so it is called Taylor series for f(x) at c
KEY TO THEOREM: says if a power series converges to f(x), the series must be a Taylor Series
𝑓(𝑛) (𝑐)
𝑛!
WHAT THEOREM DOES NOT SAY: that every series formed with the Taylor Coefficients an=
will
converge to f(x).
Definitions of Taylor and MacLaurin Series:
If a function f has derivatives of all orders at x=c, then the series:
∑∞
𝒏=𝟎
𝒇(𝒏) (𝒄)
(x-c)n
𝒏!
= f(c) + f’(c)(x-c) + … +
𝒇(𝒏) (𝒄)
𝒏!
(x-c)n +..
Moreover, if c=0, then the series is the MacLaurin Series for f
is called the Taylor Series for f(x)at c.
**If you know the pattern for the coefficients of the Taylor Polynomials for a function, you can extend
the pattern easily to form the corresponding Taylor Series.
1
1
1
Ex. P4(x) of ln x centered at 1 = (x-1) - 2(x-1)2 + 3(x-1)3 - 4(x-1)4
1
2
 Extend for Taylor Series for ln x centered at c = 1 (x-1) - (x-1)2 +…+
(−1)𝑛+1
(x-1)n
𝑛
EX….Use the function f(x) = sin x to form MacLaurin Series and determine the interval of convergence
*remember ∑∞
𝑛=0
= 0 + 1x + 0x2 -
𝑥3
3!
𝑓(𝑛) (𝑐)
(x-c)n
𝑛!
+ 0x4 +
𝑥𝑥
5!
so power series is ∑∞
𝑛=0
+ 0x6 -
𝑥7
7!
= x-
𝑥3
3!
+
𝑥5
5!
-
𝑥7
7!
𝑓(𝑛) (0) n
(x)
𝑛!
= f(0) + f’(0)x +
𝑓′′ (0) 2
x
2!
+
𝑓′′′ (0) 3
x
3!
+…
+…
𝑥 2𝑛+1
𝑛
By Ratio Test: ∑∞
𝑛=0(−1) (2𝑛+1)! converges for all x
** In previous example  CAN’T conclude that the power series converges to sin x for all x
 It converges to some function – but you are not sure what function it is
**Important point in dealing with Taylor and MacLaurin Series:
1) You must remember that the derivatives are being evaluated at every single point
2) So another function will agree with the value of f(n)(x) when x=c and disagree at other x-values
Figure:
**obtain same power series as above ex.
** Series converges for all x, but cannot
Converge to both f(x) and sinx for all x
--Let f have derivates of all orders in an open interval I centered at c
a) Taylor series for f may fail to converge for some x in I
b) Or, even if convergent, it may fail to have f(x) as its sum
THEOREM 9.23 CONVERGENCE OF TAYLOR SERIES
**If lim 𝑅𝑛 =0 for all x in the interval, I, then the Taylor Series for f converges and equals ∑∞
𝑛=0
𝑛→∞
c)n
𝑓(𝑛) (𝑐)
(x𝑛!
** Basically this is saying that: a Power Series formed with Taylor Coefficients a n =
𝒇(𝒏) (𝒄)
𝒏!
converges to
the function from which it was derived at precisely those values for which the remainder approaches
0 as n∞
Ex. Show that the MacLaurin Series for f(x) = sin x converges to sin x for all x
--From previous example sin x = x -
𝑥3
3!
+
𝑥5
5!
-
𝑥7
7!
𝑥 2𝑛+1
𝑛
+ …∑∞
𝑛=0(−1) (2𝑛+1)! is true for all x because
f(n+1)(x) = ±sin x or f(n+1)(x) = ±cos x.
 You know that 0≤ |𝑅𝑛 (𝑥)|=|
𝑓(𝑛+1) (𝑧)
(𝑛+1)!
|𝑥|𝑛+1
=
(𝑛+1)!
𝑛→∞
 Follows that for a fixed x lim
|𝑥|𝑛+1
𝑥 𝑛+1 | ≤ (𝑛+1)! (Taylor’s Theorem)
0
and by the squeeze Theorem, it follows that for all
x Rn(x)0 as n∞ the Maclaurin series for sin x converges to sin x for all x.
Summary : Guidelines for finding a Taylor Series:
1.
Differentiate f(x) several times and evaluate each derivative at c: f(c), f’(c), f’’(c), f’’’(c)..fn(c) 
Try to recognize a pattern in these #’s
2. Use the sequence developed in the 1st step to for the Taylor coefficients an =
determine the IOC for the resulting power series: f(c) + f’(c)(x-c) + … +
𝑓
(𝑛)
(𝑐)
𝑛!
𝑓(𝑛) (𝑐)
𝑛!
and
(x-c)n
3. Within the IOC determine whether of not the series converges to f(x)
**Direct determination of Taylor or MacLaurin coefficients using successive differentiation can be
difficult  below is an example of a shortcut using known coefficients of Taylor or MacLaurin Series
Ex. find the MacLaurin series for f(x) = sin x2
1) To find the coefficients you must calculate successive derivatives of f(x) = sin x2
f’(x) = 2x cos x2 f’’(x) = -4x2sin x2+ 2cosx2  to keep finding derivative is cumbersome so:
Alternate Method: 1) Maclaurin series for sin x found above is sin x = x -
𝑥3
3!
+
𝑥5
5!
-
𝑥7
7!
+…
2) Now because sin x2 = g(x)2 you can substitute x2 for x in the series to obtain:
Sin x2 = g(x)2 = x2 -
𝑥6
3!
+
𝑥 10
5!
-
𝑥 14
+
7!
…
Most practical way to find a Taylor or MacLaurin Series:
1) Develop power series from a basic list of elementary functions
2) From the list  you can determine power series for other functions by +/-/x/÷ and
differentiation /integration a composition with a known power series.
BINOMIAL SERIES – function of the form f(x) = (1+x)k
Ex. Find the Maclaurin series for f(x) = (1+x)k and determine its Radius of Convergence:
Successive differentiation = f’(x) =k(1+x)k-1, f’’(x)=k(k-1)(1+x)k-2, f’’’(x)=k(k-1)(k-2)(1+x)k-3,
fn(x)=k..(k-n+1)(1+x)k-n  f(0)=1, f’(0)=k, f’’(0)=k(k-1), f’’’(0)=k(k-1)(k-2), fn(0)=k(k-1)…(k-n+1)
Which produces the series: 1+kx +
Because
𝑎𝑛+1

𝑎𝑛
𝑘(𝑘−1)𝑥 2
2
+ …+
𝑘(𝑘−1)…(𝑘−𝑛+1)𝑥 𝑛
𝑛!
1 you can apply the ratio test to conclude the ROC is R=1, so the series converges to
some function in the interval (-1,1)
3
Ex. Find the power series for f(x) = √1 + 𝑥  (1+x)1/3
𝑘(𝑘−1)𝑥 2
𝑘(𝑘−1)…(𝑘−𝑛+1)𝑥 𝑛
+ …+
2
𝑛!
2·5𝑥 3 2·5·8𝑥 4
- 34 4! + … which converges
33 3!
1) Use the binomial series: (1+x)k = 1+kx +
2𝑥 2
2) Let k = 1/3 and write: 1+1/3x + 32 2 +
for -1≤ x ≤ 1
***Know deriving Power Series from Elementary Function chart in
book ***
Ex.. Deriving a power series from a Basic List: Find the power series for f(x) = cos√𝑥
1) Use power series for cos x = 1 2
cos√𝑥 = 1 -
√𝑥
2!
4
+
√𝑥
4!
𝑥2
2!
+
𝑥4
4!
-
𝑥6
6!
+ … and replace x by √𝑥 to get:
𝑥
2!
+
𝑥2
4!
-
𝑥3
6!
…
6
-
√𝑥
6!
= 1-
this series converges for all x in the domain of cos√𝑥  that is for all x≥0
Multiplication/Division of a power Series:
Ex. find the first 3 nonzero terms in each MacLaurin Series a. 𝑒 𝑥 arctanx
b. tan x