2-1
MAA 102 Chapter 2: Series
2.1 SERIES
Given a sequence { }na , we define a new sequence { }nS by 1 2
1
n
n k nk
S a a a a=
= = + + +
where nS is called the nth partial sum.
1 1
2 1 2
3 1 2 3
1 2 3n n
S aS a aS a a a
S a a a a
== += + +
= + + + +#
This new sequence { }nS is called the sequence of partial sums.
EXAMPLE 2.1. Derive the partial sum sequence for the following series.
(a) 11
12nn
= (b)
1
3nn
=
SOLUTION. (a)
2-2
(b)
2-3
DEFINITION 2.1. An infinite series 1 21
nn
a a a
=+ + = , with the partial sum sequence
{ }nS , converges if lim nn S exists.
If lim nn S A = , then we say that 1 nn a
= converges to A, that is Aa
nn =
=1. If lim nn S does not
exist, then the series diverges.
EXAMPLE 2.2. Determine whether the following series converges or diverges.
(a) 11
12nn
= (b)
13n
n
=
SOLUTION.
(a) 11
12nn
=
(b) 13n
n
=
2-4
EXAMPLE 2.3. Prove that the infinite series 4 8 163 23 9 27
+ + converges and find its sum.
SOLUTION.
THEOREM 2.1. (Geometric Series)
The geometric series 1 2 11
n n
nar a ar ar ar
=
= + + + + + with 0a (i) converges and has the sum
1a
r if 1r < .
(ii) diverges if 1r .
2-5
EXAMPLE 2.4. Determine whether the series 0
1 37 2
n
n
=
converges or diverges.
SOLUTION.
The converse is false, that is, if lim 0nn a = , it does not necessarily follow that the series
1n
na
= is convergent. For example, 1lim 0n n = but 1
1n n
= is divergent.
COROLLARY 2.1. (Test for Divergence)
If lim 0nn a , then the infinite series 1 nn a
= is divergent.
THEOREM 2.2. If an infinite series 1
nn
a
= is convergent, then lim 0nn a = .
2-6
EXAMPLE 2.5. Determine whether the following series converges or diverges.
(a) 1 2 3 ...3 5 7 2 1
nn
+ + + + ++ (b) ( )1 1n
n
=
SOLUTION. (a)
(b) ( )1
1 nn
= Three ways to check!
(i) Theorem 2.2
(ii) Theorem 2.1
(iii) Definition 2.1
2-7
Note that if 0 , then 1
nn
a
= and
1n
na
= will converge/diverge together.
EXAMPLE 2.6. Determine the convergence or divergence of ( )1
2 2 nnn
a =
+ if
12.75n
na
== .
SOLUTION.
THEOREM 2.3. If the convergent series 1
nn
a A
== and
1n
nb B
== , then
( )1
n nn
a b A B =
+ = + where , .
2-8
The converse of Theorem 2.3 is false, that is, if ( )1
n nn
a b =
+ is convergent, it does not necessarily follow that
1n
na
= and
1n
nb
= is convergent.
EXAMPLE 2.7. Determine the convergence or divergence of 1
1 11n n n
=
+ .
SOLUTION.
2-9
2.2 POSITIVE-TERM SERIES It is not easy to find the sum of an infinite series. We can, however, develop techniques for
using the nth term na to test a series for convergence or divergence. When applying these
tests, we shall be concerned not with the sum of the series but whether the series converges or
diverges. In this section, we shall consider only positive term series, that is, infinite series
for which every term is positive.
DEFINITION 2.2. ( ) ( )1 1
lim b
bf x dx f x dx
=
TEST 2.1. (Integral Test)
If the function ( )f x is continuous, positive valued ( )( )0f x > and decreasing for 1x , then the infinite series ( ) ( ) ( ) ( )1 2 3f f f f n+ + + + +
(i) converges if ( )1
f x dx converges.
(ii) diverges if ( )1
f x dx diverges.
2-10
EXAMPLE 2.8. Use the Integral Test to determine whether the series ( )2
231 2kk
k
= + converges or diverges.
SOLUTION.
2-11
EXAMPLE 2.9. Use the Integral Test to prove that the harmonic series 1
1n n
= diverges.
SOLUTION.
EXAMPLE 2.10. Determine whether the following series converge or diverge.
(a) 2 2 21 1 112 3 n
+ + + + + (b) 5 5 552 3 n
+ + + + +
SOLUTION. (a)
(b)
THEOREM 2.4. The p-series 1
1p
n n
=
(i) converges if 1p > . (ii) diverges if 1p .
2-12
EXAMPLE 2.11. Determine whether the following series converge or diverge.
(a) 1
14 3nn
= + (b) 212
n
nn
=
+
SOLUTION. (a)
(b)
TEST 2.2. (Comparison Test)
Suppose 0 n na b for every positive integer n.
(i) If 1
nn
b
= converges, then
1n
na
= converges.
(ii) If 1
nn
a
= diverges, then
1n
nb
= diverges.
2-13
LIMIT COMPARISON TEST
Suppose 1
nn
a
= and
1n
nb
= are positive term series with lim nn
n
a Ab
= .
(i) If 0 A< < , then 1
nn
a
= and
1n
nb
= both converge or both diverge.
(ii) If 0A = and 1
nn
b
= converges, then
1n
na
= converges.
(iii) If A = and 1
nn
b
= diverges, then
1n
na
= diverges.
USING THE COMPARISON TEST STEP 1: Guess at whether the series na converges or diverges. STEP 2: Find a series that proves the guess to be correct. That is, if we guess that na diverges, we must find a divergent series whose
terms are smaller than the corresponding terms of na . If we guess that na converges, we must find a convergent series whose terms
are larger than the corresponding term of na . Some informal principles can be used to find this divergent or convergent series.
1. Constant terms in the denominator of na can usually be deleted without affecting the convergence or divergence of the series.
2. If a polynomial in n appears as a factor in the numerator or denominator of na , all but the leading term in the polynomial can usually be discarded without affecting the convergence or divergence of the series.
Unfortunately, it is not always so straightforward to find the series required for comparison, so the Limit Comparison Test is introduced as an alternative to the comparison test.
2-14
EXAMPLE 2.12. Determine whether the following series converge or diverge.
(a) 3 2
1
12n n
= + (b) 2121n
nn
=
++
SOLUTION.
(a)
(b)
2-15
2.3 ALTERNATING SERIES
An infinite series whose terms are alternately positive and negative is called an alternating
series. It is customary to express an alternating series in one of the forms
( ) 11 2 3 4 1 n na a a a a + + + + or
( )1 2 3 4 1 n na a a a a + + + +
If an alternating series satisfies the hypotheses of the alternating series test, and if S is the
sum of the series, then:
(a) S lies between any two successive partial sums; that is, either
1n nS S S + or 1n nS S S+ depending on which partial sum is larger.
(b) If S is approximated by Sn, then the absolute error nS S satisfies 1n nS S a + . Moreover, the sign of the error nS S is the same as that of the coefficient of 1na + .
TEST 2.3. (Alternating Series Test)
Suppose an alternating series ( ) 11
1 n nn
a +
= fulfils the following conditions:
(i) 0na > , n (ii) 1n na a+ < , n (iii) lim 0nn a = .
Then the series is convergent.
2-16
EXAMPLE 2.13. Determine whether the following alternating series converge or diverge.
(a) ( ) 11
11 nn n
+
= (b) ( ) 1 2
1
214 3
n
n
nn
=
SOLUTION.
(a)
(b)
2-17
DEFINITION 2.3. An infinite series 1
nn
a
= is absolutely convergent if the series
1 21
n nn
a a a a
== + + + +
obtained by taking the absolute value of each term is convergent. The series is said to be
absolutely divergent if the series of absolute values diverges.
Note that if 1
nn
a
= is a positive term series, then n na a= , and in this case absolute
convergence is the same as convergence.
EXAMPLE 2.14. Prove that the alternating series ( ) 121
1 n
n n
+
=
is absolutely convergent.
SOLUTION.
The converse of Theorem 2.5 is false, that is, if 1
nn
a
= is convergent, it does not necessarily
follow that 1
nn
a
= is convergent.
THEOREM 2.5. If an infinite series 1
nn
a
= is absolutely convergent, then
1n
na
= is
convergent.
2-18
DEFINITION 2.4. An infinite series 1
nn
a
= is conditionally convergent if
1n
na
= is
convergent and 1
nn
a
= is divergent.
EXAMPLE 2.15. Show that the alternating series ( )1
1 n
n n
=
is conditionally convergent.
SOLUTION.
2-19
2.4 RATIO TEST
TEST 2.4. (Ratio Test)
Let 1
nn
a
= be an infinite series of nonzero terms. If 1lim nn
n
a La+
= , then 1 nn a
=
(i) is absolutely convergent if 1L < . (ii) is divergent if 1L > or L = . (iii) may be absolutely convergent, conditionally convergent or divergent if 1L = .
2-20
EXAMPLE 2.16. Determine whether the following series are absolutely convergent,
conditionally convergent or divergent.
(a) 1
1!n n
= (b) ( )
1
!1 n nn
ne
=
(c) 1
1n n
=
SOLUTION. (a)
(b)
(c)
2-21
2.5 ROOT TEST
EXAMPLE 2.17. Determine the convergence or divergence of the following series.
(a) 11 2nn
n
= (b)
1
23
nn
nn
=
SOLUTION.
(a)
(b)
TEST 2.5. (Root Test)
Let 1
nn
a
= be an infinite series. If Lan nn =lim , then 1 nn a
=
(i) is absolutely convergent if 1L < . (ii) is divergent if 1L > or L = . (iii) may be absolutely convergent, conditionally convergent or divergent if 1L = .
2-22
SUMMARY
To determine the convergence of an infinite series1
nn
a
= :
STEP 1 : Find lim nn a .
(i) If lim 0nn a , then 1 nn a
= is divergent (Collorary 2.1).
(ii) If lim 0nn a = , then no conclusion can be made. Go to Step 2.
STEP 2 : Determine whether 1
nn
a
= is positive term series or alternating series.
(i) If positive term series, use
integral test comparison test ratio test root test
(ii) If alternating series, use
alternating series test ratio test root test
Ratio Test and Root Test can be used for any series, namely positive term series and
alternating series.
Ratio Test : Series with factorials (!) and (expression)n.
Root Test : Series with (expression)n
2-23
STRATEGY FOR SERIES
STEP 1: With a quick glance does it look like the series terms dont converge to zero in the
limit, i.e. does lim 0nn a ? If so, use the Divergence Test. Note that you should only do the divergence test if a quick glance suggests that the series terms may
not converge to zero in the limit.
STEP 2: Is the series a p-series ( 1pn ) or a geometric series ( 1 nn ar
= or 1
1
n
nar
= )? If so
use the fact that p-series will only converge if 1p > and a geometric series will only converge if 1r < . Remember as well that often some algebraic manipulation is required to get a geometric series into the correct form.
STEP 3: Is the series similar to a p-series or a geometric series? If so, try the Comparison
Test.
STEP 4: Is the series a rational expression involving only polynomials or polynomials under
radicals (i.e. a fraction involving only polynomials or polynomials under radicals)?
If so, try the Comparison Test and/or the Limit Comparison Test. Remember
however, that in order to use the Comparison Test and the Limit Comparison Test
the series terms all need to be positive.
STEP 5: Does the series contain factorials or constants raised to powers involving n? If so,
then the Ratio Test may work. Note that if the series term contains a factorial then
the only test that weve got that will work is the Ratio Test.
STEP 6: Can the series terms be written in the form ( )1 nn na b= or ( ) 11 nn na b+= ? If so, then the Alternating Series Test may work.
STEP 7: Can the series terms be written in the form ( )nn na b= ? If so, then the Root Test may work.
STEP 8: If ( )na f n= for some positive, decreasing function and ( )a f x dx is easy to evaluate then the Integral Test may work.
2-24
EXTRA NOTES
Divergence Test
If lim 0nn a then na will diverge
Integral Test
Suppose that ( )f x is a positive, decreasing function on the interval [ ),k and that ( ) nf n a= then,
1. If ( ) k
f x dx is convergent so is n
n ka
= .
2. If ( ) k
f x dx is divergent so is n
n ka
= .
Comparison Test
Suppose that we have two series na and nb with , 0n na b for all n and n na b for all n. Then,
1. If nb is convergent then so is na . 2. If na is divergent then so is nb .
Limit Comparison Test
Suppose that we have two series na and nb with , 0n na b for all n. Define, lim n
nn
acb
= .
If c is positive (i.e. 0c > ) and is finite (i.e. c < ) then either both series converge or both series diverge.
2-25
Alternating Series Test
Suppose that we have a series na and either ( )1 nn na b= or ( ) 11 nn na b+= where 0nb for all n. Then if,
1. lim 0nn b = and, 2. { }nb is eventually a decreasing sequence,
Then the series is convergent.
Ratio Test
Suppose we have the series na . Define, 1lim n
nn
aLa+
=
Then,
1. if 1L < the series is absolutely convergent (and hence convergent). 2. if 1L > the series is divergent. 3. if 1L = the series may be divergent, conditionally convergent, or absolutely
convergent.
Root Test
Suppose that we have the series na . Define,
1
lim limn nn nn nL a a = = Then,
1. if 1L < the series is absolutely convergent (and hence convergent). 2. if 1L > the series is divergent. 3. if 1L = the series may be divergent, conditionally convergent, or absolutely
convergent.