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11 INFINITE SEQUENCES AND SERIES
11INFINITE SEQUENCES AND SERIES
11.2Series
In this section, we will learn about:
Various types of series.
INFINITE SEQUENCES AND SERIES
SERIES
If we try to add the terms of an infinite sequence we get an expression of the form
a1 + a2 + a3 + ··· + an + ∙·∙
1{ }n na ∞=
Series 1
INFINITE SERIES
This is called an infinite series (or just a series).
It is denoted, for short, by the symbol
1orn n
na a
∞
=∑ ∑
However, does it make sense to talk about the sum of infinitely many terms?
INFINITE SERIES
It would be impossible to find a finite sum for the series
1 + 2 + 3 + 4 + 5 + ∙∙∙ + n + ···
If we start adding the terms, we get the cumulative sums 1, 3, 6, 10, 15, 21, . . .
After the nth term, we get n(n + 1)/2, which becomes very large as n increases.
INFINITE SERIES
However, if we start to add the terms of the series
we get:
1 1 1 1 1 1 12 4 8 16 32 64 2n+ + + + + + +⋅⋅⋅ + ⋅⋅⋅
3 7 15 31 6312 4 8 16 32 64, , , , , , ,1 1/ 2 ,n⋅ ⋅ −⋅ ⋅⋅ ⋅
INFINITE SERIES
The table shows that, as we add more and more terms, these partial sums become closer and closer to 1.
In fact, by adding sufficiently many terms of the series, we can make the partial sums as close as we like to 1.
INFINITE SERIES
So, it seems reasonable to say that the sum of this infinite series is 1 and to write:
1
1 1 1 1 1 1 12 2 4 8 16 2n n
n
∞
=
= + + + + + +⋅⋅⋅ ⋅ ⋅ ⋅ =∑
INFINITE SERIES
We use a similar idea to determine whether or not a general series (Series 1) has a sum.
INFINITE SERIES
We consider the partial sums
s1 = a1
s2 = a1 + a2
s3 = a1 + a2 + a3
s3 = a1 + a2 + a3 + a4
In general, 1 2 31
n
n n ii
s a a a a a=
⋅⋅ + =⋅= + + + ∑
INFINITE SERIES
These partial sums form a new sequence {sn}, which may or may not have a limit.
INFINITE SERIES
If exists (as a finite number), then, as in the preceding example, we call it the sum of the infinite series Σ an.
lim nns s
→∞=
SUM OF INFINITE SERIES
Given a series
let sn denote its nth partial sum:
1 2 31n
na a a a∞
=
= + + + ⋅⋅⋅∑
1 21
n
n i ni
s a a a a=
= = + + ⋅⋅⋅+∑
SUM OF INFINITE SERIES Definition 2
If the sequence {sn} is convergent and exists as a real number, then the
series Σ an is called convergent and we write:
The number s is called the sum of the series.
Otherwise, the series is called divergent.
1 21
orn nn
a a a s a s∞
=
+ + ⋅⋅⋅ + + ⋅⋅⋅ = =∑
SUM OF INFINITE SERIES Definition 2
lim nns s
→∞=
Thus, the sum of a series is the limit of the sequence of partial sums.
So, when we write ,
we mean that, by adding sufficiently many terms of the series, we can get as close as we like to the number s.
1n
na s
∞
=
=∑
SUM OF INFINITE SERIES
Notice that:
1 1lim
n
n inn ia a
∞
→∞= =
=∑ ∑
SUM OF INFINITE SERIES
Compare with the improper integral
To find this integral, we integrate from 1 to tand then let t → ∞.
For a series, we sum from 1 to n and then let n → ∞.
1 1( ) lim ( )
t
tf x dx f x dx
∞
→∞=∫ ∫
SUM OF INFINITE SERIES VS. IMPROPER INTEGRALS
An important example of an infinite series is the geometric series
2 3 1
1
10
n
n
n
a ar ar ar ar
ar a
−
∞−
=
+ + + + ⋅⋅⋅+ + ⋅⋅⋅
= ≠∑
GEOMETRIC SERIES Example 1
Each term is obtained from the preceding one by multiplying it by the common ratio r.
We have already considered the special case where a = ½ and r = ½ earlier in the section.
GEOMETRIC SERIES Example 1
If r = 1, then
sn = a + a + ∙∙∙ + a = na → ±∞
Since doesn’t exist, the geometric series diverges in this case.
GEOMETRIC SERIES Example 1
lim nns
→∞
If r ≠ 1, we have
sn = a + ar + ar2 + ∙∙∙ + ar n–1
and
rsn = ar + ar2 + ∙∙∙ +ar n–1 + ar n
GEOMETRIC SERIES Example 1
Subtracting these equations, we get:
sn – rsn = a – ar n
(1 )1
n
na rs
r−
=−
GEOMETRIC SERIES E. g. 1—Equation 3
If –1 < r < 1, we know from Result 9 in Section 11.1 that r n → 0 as n → ∞.
So,
Thus, when |r | < 1, the series is convergent and its sum is a/(1 – r).
(1 )lim lim lim1 1 1 1
nn
nn n n
a r a a as rr r r r→∞ →∞ →∞
−= = − =
− − − −
GEOMETRIC SERIES Example 1
If r ≤ –1 or r > 1, the sequence {r n} is divergent by Result 9 in Section 11.1
So, by Equation 3, does not exist.
Hence, the series diverges in those cases.
GEOMETRIC SERIES Example 1
lim nns
→∞
The figure provides a geometric demonstration of the result in Example 1.
GEOMETRIC SERIES
If s is the sum of the series, then, by similar triangles,
So,
s aa a ar=
−
GEOMETRIC SERIES
1as
r=
−
GEOMETRIC SERIES
We summarize the results of Example 1 as follows.
The geometric series
is convergent if |r | < 1.
1 2
1
n
nar a ar ar
∞−
=
= + + + ⋅⋅⋅∑
GEOMETRIC SERIES Result 4
The sum of the series is:
If |r | ≥ 1, the series is divergent.
1
11
1n
n
aar rr
∞−
=
= <−∑
GEOMETRIC SERIES Result 4
Find the sum of the geometric series
The first term is a = 5 and the common ratio is r = –2/3
10 20 403 9 275− + − + ⋅⋅⋅
GEOMETRIC SERIES Example 2
Since |r | = 2/3 < 1, the series is convergent by Result 4 and its sum is:
23
53
10 20 40 553 9 27 1 ( )
5
3
⋅ ⋅− + − +−
=
⋅ =−
=
GEOMETRIC SERIES Example 2
What do we really mean when we say that the sum of the series in Example 2 is 3?
Of course, we can’t literally add an infinite number of terms, one by one.
GEOMETRIC SERIES
However, according to Definition 2, the total sum is the limit of the sequence of partial sums.
So, by taking the sum of sufficiently many terms, we can get as close as we like to the number 3.
GEOMETRIC SERIES
The table shows the first ten partial sums sn.
The graph shows how the sequence of partial sums approaches 3.
GEOMETRIC SERIES
Is the series
convergent or divergent?
2 1
12 3n n
n
∞−
=∑
GEOMETRIC SERIES Example 3
Let’s rewrite the nth term of the series in the form ar n-1:
We recognize this series as a geometric series with a = 4 and r = 4/3.
Since r > 1, the series diverges by Result 4.
12 1 2 ( 1)
11 1 1 1
4 42 3 (2 ) 3 43 3
nnn n n n
nn n n n
−∞ ∞ ∞ ∞− − −
−= = = =
= = =
∑ ∑ ∑ ∑
GEOMETRIC SERIES Example 3
Write the number as a ratio of integers.
2.3171717…
After the first term, we have a geometric series with a = 17/103 and r = 1/102.
3 5 7
17 17 172.310 10 10
= + + + + ⋅⋅⋅
GEOMETRIC SERIES Example 4
2.317 2.3171717...=
Therefore,
3
2
17 1710 10002.317 2.3 2.31 991
10 10023 1710 9901147495
= + = +−
= +
=
GEOMETRIC SERIES Example 4
Find the sum of the series where |x| < 1.
Notice that this series starts with n = 0.
So, the first term is x0 = 1.
With series, we adopt the convention that x0 = 1 even when x = 0.
0
n
nx
∞
=∑
GEOMETRIC SERIES Example 5
Thus,
This is a geometric series with a = 1 and r = x.
2 3 4
01n
nx x x x x
∞
=
= + + + + + ⋅⋅⋅∑
GEOMETRIC SERIES Example 5
Since |r | = |x| < 1, it converges, and Result 4 gives:
0
11
n
nx
x
∞
=
=−∑
GEOMETRIC SERIES E. g. 5—Equation 5
Show that the series
is convergent, and find its sum.
1
1( 1)n n n
∞
= +∑
SERIES Example 6
This is not a geometric series.
So, we go back to the definition of a convergent series and compute the partial sums:
1
1( 1)
1 1 1 11 2 2 3 3 4 ( 1)
n
ni
si i
n n
=
=+
= + + + +⋅ ⋅ ⋅
⋅ ⋅+
⋅
∑
SERIES Example 6
We can simplify this expression if we use the partial fraction decomposition.
See Section 7.4
1 1 1( 1) 1i i i i
= −+ +
SERIES Example 6
Thus, we have:
1
1
1( 1)1 1
11 1 1 1 1 1 112 2 3 3 4 111
1
n
ni
n
i
si i
i i
n n
n
=
=
=+
= − + = − + − + − + ⋅⋅⋅+ − +
= −+
∑
∑
SERIES Example 6
Thus,
Hence, the given series is convergentand
1lim lim 1 1 0 11nn n
sn→∞ →∞
= − = − = +
1
1 1( 1)n n n
∞
=
=+∑
SERIES Example 6
The figure illustrates Example 6 by showing the graphs of the sequence of terms an =1/[n(n + 1)] and the sequence {sn} of partial sums.
Notice that an → 0 and sn → 1.
SERIES
Show that the harmonic series
is divergent.
1
1 1 1 112 3 4n n
∞
=
= + + + + ⋅⋅⋅∑
HARMONIC SERIES Example 7
For this particular series it’s convenient to consider the partial sums s2, s4, s8, s16, s32, …and show that they become large.
( ) ( )
11
2 2
1 1 1 1 1 14 2 3 4 2 4 4
22
1
1
1 1
1
s
s
s >
=
= +
= + + + + + +
= +
HARMONIC SERIES Example 7
( )( ) ( )
1 1 1 1 1 1 18 2 3 4 5 6 7 8
1 1 1 1 1 1 12 4 4 8 8 8 8
1 1 12 2 232
1
1
1
1
s
= + + + + + + +
> + + + + + + +
= + + +
= +
HARMONIC SERIES Example 7
Similarly,
( ) ( )( ) ( ) ( )
1 1 1 1 1 1 116 2 3 4 5 8 9 16
1 1 1 1 1 1 12 4 4 8 8 16 16
1 1 1 12 2 2 242
1
1
1
1
s
= + + + + + ⋅⋅⋅+ + + ⋅⋅⋅ +
> + + + + + ⋅⋅⋅+ + + ⋅⋅⋅ +
= + + + +
= +
HARMONIC SERIES Example 7
Similarly,
Similarly, s32 > 1 + 5/2, s64 > 1 + 6/2, and, in general,
This shows that s2n → ∞ as n → ∞,and so {sn} is divergent. Therefore, the harmonic series diverges.
HARMONIC SERIES Example 7
21
2n
ns > +
HARMONIC SERIES
The method used in Example 7 for showing that the harmonic series diverges is due to the French scholar Nicole Oresme (1323–1382).
If the series is convergent,
then
1n
na
∞
=∑
lim 0nna
→∞=
SERIES Theorem 6
Let sn = a1 + a2 + ∙∙∙ + an
Then, an = sn – sn–1
Since Σ an is convergent, the sequence {sn} is convergent.
SERIES Theorem 6—Proof
Let
Since n – 1 → ∞ as n → ∞, we also have:
SERIES Theorem 6—Proof
lim nns s
→∞=
1lim nns s−→∞
=
Therefore,
( )1
1
lim lim
lim lim
0
n n nn n
n nn n
a s s
s s
s s
−→∞ →∞
−→∞ →∞
= −
= −
= −=
SERIES Theorem 6—Proof
With any series Σ an we associate two sequences:
The sequence {sn} of its partial sums
The sequence {an} of its terms
SERIES Note 1
If Σ an is convergent, then
The limit of the sequence {sn} is s (the sum of the series).
The limit of the sequence {an}, as Theorem 6 asserts, is 0.
SERIES Note 1
The converse of Theorem 6 is not true in general.
If , we cannot conclude
that Σ an is convergent.
SERIES Note 2
lim 0nna
→∞=
Observe that, for the harmonic series Σ 1/n,we have an = 1/n → 0 as n → ∞.
However, we showed in Example 7 that Σ 1/nis divergent.
SERIES Note 2
If does not exist or if ,
then the series
is divergent.
lim nna
→∞lim 0nn
a→∞
≠
1n
na
∞
=∑
THE TEST FOR DIVERGENCE Test 7
The Test for Divergence follows from Theorem 6.
If the series is not divergent, then it is convergent.
Thus,
TEST FOR DIVERGENCE
lim 0nna
→∞=
Show that the series diverges.
=
So, the series diverges by the Test for Divergence.
2
21 5 4n
nn
∞
= +∑TEST FOR DIVERGENCE Example 8
2
2 2
1 1lim lim lim 05 4 5 4 / 5nn n n
nan n→∞ →∞ →∞
= = ≠+ +
If we find that , we know that Σ an
is divergent.
If we find that , we know nothingabout the convergence or divergence of Σ an.
SERIES Note 3
lim 0nna
→∞≠
lim 0nna
→∞=
Remember the warning in Note 2:
If , the series Σ an might converge or diverge.
SERIES
lim 0nna
→∞=
Note 3
If Σ an and Σ bn are convergent series, then so are the series Σ can (where c is a constant), Σ (an + bn), and Σ (an – bn), and
( )
( )
1 1
1 1 1
1 1 1
i.
ii.
iii.
n nn n
n n n nn n n
n n n nn n n
ca c a
a b a b
a b a b
∞ ∞
= =
∞ ∞ ∞
= = =
∞ ∞ ∞
= = =
=
+ = +
− = −
∑ ∑
∑ ∑ ∑
∑ ∑ ∑
SERIES Theorem 8
These properties of convergent series follow from the corresponding Limit Laws for Sequences in Section 11.1
For instance, we prove part ii of Theorem 8 as follows.
SERIES
Let
1 1
1 1
n
n i ni n
n
n i ni n
s a s a
t b t b
∞
= =
∞
= =
= =
= =
∑ ∑
∑ ∑
THEOREM 8 ii—PROOF
The nth partial sum for the series Σ (an + bn) is:
( )1
n
n iii
u a b=
= +∑
THEOREM 8 ii—PROOF
Using Equation 10 in Section 5.2, we have:
( )1
1 1
1 1
lim lim
lim
lim lim
lim lim
n
n i in n i
n n
i in i i
n n
i in ni i
n nn n
u a b
a b
a b
s t s t
→∞ →∞=
→∞= =
→∞ →∞= =
→∞ →∞
= +
= +
= +
= + = +
∑
∑ ∑
∑ ∑
THEOREM 8 ii—PROOF
Hence, Σ (an + bn) is convergent, and its sum is:
( )1
1 1
n nn
n nn n
a b s t
a b
∞
=
∞ ∞
= =
+ = +
= +
∑
∑ ∑
THEOREM 8 ii—PROOF
Find the sum of the series
The series Σ 1/2n is a geometric series with a = ½ and r = ½. Hence,
1
3 1( 1) 2n
n n n
∞
=
+ +
∑
SERIES Example 9
12
11 2
1 12 1n
n
∞
=
= =−∑
In Example 6, we found that:
So, by Theorem 8, the given series is convergent and
1 1 1
3 1 1 13( 1) 2 ( 1) 2
3 1 14
n nn n nn n n n
∞ ∞ ∞
= = =
+ = + + +
= ⋅ +=
∑ ∑ ∑
SERIES Example 9
1
1 1( 1)n n n
∞
=
=+∑
A finite number of terms doesn’t affect the convergence or divergence of a series.
SERIES Note 4
For instance, suppose that we were able to show that the series is convergent.
Since
it follows that the entire series is convergent.
34 1n
nn
∞
= +∑
SERIES Note 4
3 31 4
1 2 31 2 9 28 1n n
n nn n
∞ ∞
= =
= + + ++ +∑ ∑
31 1n
nn
∞
= +∑
Similarly, if it is known that the series converges, then the full series
is also convergent.
1n
n Na
∞
= +∑
1 1 1
N
n n nn n n N
a a a∞ ∞
= = = +
= +∑ ∑ ∑
SERIES Note 4
