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Math 152, Benjamin
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10.1 Sequences
A sequence is an ordered list of numbers: a1 , a2 , a3 , . . . , an , an+1 , . . .
Each of the numbers is called a term of the sequence.
Notation: A sequence {a1 , a2 , a3 , . . .} can be denoted by {an } or {an }∞
n=1 . Unless otherwise stated, we
assume n starts at 1, but this does not always have to be the case.
Examples: Find the first 4 terms of the following sequences.
•
(
n2
n+4
)∞
n=1
• an = (−1)n n3
The above sequence is called an alternating sequence since the terms alternate signs.
Find a general formula an for the sequences below.
•
•
3 4 5 6
, , , ,...
4 9 16 25
2 4 8 16
− , ,− , ,...
3 5 7 9
• {1, −2, 6, −24, 120, . . .}
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Math 152, Benjamin
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The limit of a sequence is defined to be lim an .
n→∞
If the limit exists to a finite number L, i.e, lim an = L, we say the sequence converges to L. If the limit
n→∞
does not exist or is infinite, we say the sequence diverges.
Determine whether the following sequences converge or diverge. If the sequence converges, state the limit.
• an = 2 +
• an = e
n
1
3
n2 +5
n+3
• an = arctan
• an =
−n3 + 3
√
2n + 1
!
n
, n≥2
ln n
• an = cos
3n + 7
n2 + n
2
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Math 152, Benjamin
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• an =
1
2
ln(3 + 7n3 ) − 12 ln(n3 + 2n)
• an = n −
p
n2 + 1
• an = (−1)n + 3
• an = sin(nπ)
• an = cos(nπ)
For alternating sequences, use the following important fact: If lim |an | = 0, then lim an = 0.
n→∞
n→∞
In other words, an alternating sequence converges if and only if the absolute value of the terms goes to 0 as
n goes to ∞.
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Math 152, Benjamin
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• an =
cos(nπ)
n
• an = (−3)n
• an = (−1)n
n
1
4
• an =
(−1)n+1 (3n − 1)
n4 + 3
• an =
(−1)n (n + 1)
5n + 2
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Math 152, Benjamin
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A sequence {an } is said to be increasing if an < an+1 for all n ≥ 1.
A sequence {an } is said to be decreasing if an > an+1 for all n ≥ 1.
If a sequence is either an increasing or a decreasing sequence, we say it is monotonic.
A sequence {an } is bounded if there are numbers m and M such that m ≤ an ≤ M for all n ≥ 1. We say
that {an } is bounded above by M and bounded below by m.
Theorem: Every bounded, monotonic sequence converges.
Determine whether the following sequences are increasing, decreasing, or not monotonic. Also state if the
sequence is bounded or not.
• an =
1
2 + 5n
• an =
(−1)n
2 + 5n
• an = n −
1
n
• an = e1/n
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Math 152, Benjamin
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nπ
• an = sin
2
• an =
2n
n+1
A recursive sequence is one in which terms are defined by using previous terms in the sequence. In a
recursive sequence, you have to be given at least the first term.
Find the next 4 terms of the recursively defined sequences. Do these sequences converge?
an
a1 = 1, an+1 = 4 − an
a1 = 3, an+1 = 2 +
3
The sequence below is bounded and decreasing. Find the next two terms of the sequence and find the limit.
a1 = 4, an+1 =
12
8 − an
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Math 152, Benjamin
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10.2 Series
A series is the sum of an infinite sequence of numbers. So, given a sequence {an }, the series
defined to be
∞
X
∞
P
n=1
an is
an = a1 + a2 + a3 + . . .
n=1
Can an infinite sequence of numbers have a finite sum? Sometimes yes and sometimes no! (This is similar
to improper integrals where an infinite interval may or may not have a finite area.)
If
∞
P
n=1
an = s where s is a finite number, then the series converges to s and we say s is the sum of the
series. If the limit does not exist (or is infinite), the series diverges.
Given a series
∞
P
n=1
s1 = a1
an , the nth partial sum, denoted sn is defined to be the sum of the first n terms.
s2 = a1 + a2
s3 = a1 + a2 + a3
In general, sn = a1 + a2 + a3 + . . . + an
For the following series, find the first 4 partial sums.
•
•
•
•
∞
P
1
n=1
∞
P
(−1)n
∞
P
3n
n=1
n=1
∞
X
1
n=1
n
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Math 152, Benjamin
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Consider the sequence of partial sums, {sn } = {s1 , s2 , s3 , s4 , . . .}. If this infinite sequence of partial sums
converges to a number, that number is the sum of the series. In other words, a series converges if its
sequence of partial sums converges, and
∞
X
an = lim sn
n→∞
n=1
∞
P
Example: Suppose for the series
n=1
an , it is known that sn =
3n − 1
.
2n + 6
What is the sum of the first 5 terms?
What is the sum of the first 10 terms?
Does the series converge or diverge? If it converges, what is the sum?
What is a1 ?
What is a8 ?
Find a formula for an for n > 1.
Example: Suppose for the series
∞
P
n=1
If it converges, what is the sum?
an , it is known that sn = 2+
8
n
3
2
. Does the series converge or diverge?
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Math 152, Benjamin
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In general, finding a formula for sn is not easy. However, a type of series for which a formula for sn can
often be determined are telescoping series. Telescoping series are series in which all but a finite number
of terms of the series cancel out with other terms.
Examples: Find a formula for the nth partial sum for the following series. Does the series converge? If so,
what is its sum?
∞
P
2
2
−
n+3
n=1 n + 2
n+2
ln
n+1
n=2
∞
P
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Math 152, Benjamin
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∞
P
n=3
1
1
− cos
cos
n−1
n+1
If a series is not always written in a telescoping form, partial fractions may sometimes be used to write it
in telescoping form.
∞
P
n=4
n2
−2
− 2n
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Math 152, Benjamin
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If we can find a formula for sn for a series, we can determine if the series converges or diverges, but this is
usually not easy. If you are not given a formula for sn or cannot determine one, how can you determine if
a series converges or diverges? There are many tests we will learn. The first is the Test for Divergence.
Test for Divergence: If lim an 6= 0 or does not exist, then the series
n→∞
If a series
∞
P
n=1
∞
P
n=1
an diverges.
an converges, then it first must be true that lim an = 0.
n→∞
However, if lim an = 0, this DOES NOT NECESSARILY MEAN THE SERIES CONVERGES!!!
n→∞
The classic example of a series that does NOT converge even though its terms go to 0 is the
∞ 1
∞ 1
P
P
harmonic series:
. However, the series
DOES converge!
2
n=1 n
n=1 n
What can you conclude about the following series with regards to convergence/divergence using the Test
for Divergence?
n2
n=1 n + 7
∞
P
∞
P
cos(nπ)
n=1
∞ n+3
P
n=1
n2 + 5
∞ n+9
P
n=1
∞
P
3n − 7
arctan(e−n )
n=1
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Math 152, Benjamin
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A geometric series is a series in which each term of the series is some ratio r times the previous term,
where r is some real number. These types of series can be represented as
∞
P
ar n .
n=0
It can be shown that for a geometric series of this form, sn =
a − rn
.
1−r
a − rn
converges to a finite number when −1 < r < 1, or equivalently, when |r| < 1.
n→∞
n→∞ 1 − r
a
In this case, the sum of the geometric series is
where a is the first term of the series.
1−r
lim sn = lim
Note that not all series start at n = 0. No matter what number the index starts with, a is still the first
term of the series.
Examples: For each of the following series, determine if the series converges or diverges. If it converges, find
the sum.
25 125 625
+
−
+ ···+
5−
6
36
216
∞
P
n=0
5
n
3
4
∞ 4(−1)n
P
n=1
32n
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Math 152, Benjamin
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∞
P
(−2)n+1 5−n
n=2
∞ −52n+1
P
n=1
6n−1
∞ −3 + en
P
n=0
2(7n+1 )
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Math 152, Benjamin
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For what values of x do the following series converge? For these values of x, find the sum of the series.
xn
n+1
n=1 (−8)
∞
P
∞ 2n+1 (x + 3)n
P
n=0
3(4n )
So far, the only series we can actually find a sum for are:
• Series for which we are given sn or can easily find it.
• Convergent telescoping series.
• Convergent geometric series.
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