Absolute convergence

P
Definition A series
P an is called absolutely convergent if the series of
absolute values |an | is convergent.
If the terms of the series an are positive, absolute convergence is the same as
convergence.

Example Are the following series absolutely convergent?

∞ ∞
X (−1)n+1 X (−1)n
, .
n=1
n3 n=1
n

(−1)n+1
To check if the series ∞
P
I
n=1 n3
isP
absolutely convergent, we need to
check if the series of absolute values ∞ 1
n=1 n3 is convergent.
P∞ 1
I Since n=1 3 is a p-series with p = 3 > 1, it converges and therefore
P∞ (−1)n+1n
n=1 n3
is absolutely convergent.
(−1)n
To check if the series ∞
P
I
n=1 n
is absolutely convergent, we need to
check if the series of absolute values ∞ 1
P
n=1 n is convergent.
P∞ 1
I Since n=1 n is a p-series with p = 1, it diverges and therefore
P∞ (−1)n
n=1 n
is not absolutely convergent.
Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test
Conditional convergence

P
Definition A series an is called conditionally convergent if the series is
convergent but not absolutely convergent.
Which of the series in the above example is conditionally convergent?
P∞ (−1)n+1
I Since the series is absolutely convergent, it is not
n=1 n3
conditionally convergent.
P∞ (−1)n
I Since the series is convergent (used the alternating series test
n=1 n
last day to show this), but the series of absolute values ∞ 1
P
n=1 n is not
(−1)n
convergent, the series ∞
P
n=1 n
is conditionally convergent.

Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test

Absolute conv. implies conv.
Theorem
P If a series is absolutely
P convergent, then it is convergent, that is
if |an | is convergent, then an is convergent.
(A proof is given in your notes)
Example Are the following series convergent (test for absolute convergence)
∞ ∞
X (−1)n+1 X sin(n)
, .
n=1
n3 n=1
n4

(−1)n+1
Since ∞
P
I
n=1 n3
is absolutely convergent, we can conclude that this
series is convergent.
To check if the series ∞ sin(n)
P
n=1 n4˛ is absolutely convergent, we consider the
I
P∞ ˛ sin(n) ˛˛
series of absolute values n=1 ˛ n4 ˛.
˛ ˛
I Since 0 ≤ | sin(n)| ≤ 1, we have 0 ≤ ˛ sin(n) ≤ n14 .
˛ ˛
n4 ˛
˛ ˛
Therefore the series ∞
P ˛ sin(n) ˛
n=1 ˛ n4 ˛ converges by comparison with the
I

converging p-series ∞ 1
P
n=1 n4 .
P∞ sin(n)
I Therefore the series n=1 n4 is convergent since it is absolutely
convergent.
Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test
The Ratio Test
This test is useful for determining absolute convergence.
Let ∞
P
n=1 an be a˛ series
˛ (the terms may be positive or negative).
˛a ˛
Let L = limn→∞ ˛ n+1 an ˛.
P∞
I If L < 1, then the series
n=1 an converges absolutely (and hence is
convergent).
P∞
I If L > 1 or ∞, then the series
n=1 an is divergent.
I If L = 1, then the Ratio test is inconclusive and we cannot determine if
the series converges or diverges using this test.
This test is especially useful where factorials and powers of a constant appear
in terms of a series. (Note that when the ratio test is inconclusive for an
alternating series, the alternating series test may work. )
Example 1 Test the following series for convergence
∞
X 2n
(−1)n−1
n=1
n!
˛ ˛ ˛ 2n+1 ‹(n+1)! ˛
˛ an+1 ˛ 2
I limn→∞ ˛ an ˛ = limn→∞ ˛ ˛ = limn→∞ = 0 < 1.
˛ ‹ ˛
n 2 n! n+1

I Therefore, the series converges.

Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test
Example 2

Ratio Test Let ˛ ∞

P
n=1 ˛an be a series (the terms may be positive or negative).
˛ an+1 ˛
Let L = limn→∞ ˛ an ˛.
If L < 1, then the series ∞
P
an converges absolutely.
n=1 P
If L > 1 or ∞, then the series ∞ n=1 an is divergent.
If L = 1, then the Ratio test is inconclusive.
Example 2 Test the following series for convergence
∞
X “n”
(−1)n
n=1
5n

˛ ˛ ˛ (n+1)‹5n+1 ˛
˛a ˛ n+1
I limn→∞ ˛ n+1 = lim ˛ = limn→∞ =
˛ ˛
an ˛ n→∞ ˛ ‹ 5n
n
n 5
1 1
5
limn→∞ (1 + 1/n) = 5
< 1.
I Therefore, the series converges.

Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test

Example 3

nn
Test the following series for convergence ∞
P
Example 3 n=1 n!
˛ ˛ ˛ (n+1)n+1 ‹(n+1)! ˛
˛a ˛ (n+1)(n+1)n n!
I limn→∞ ˛ n+1 = lim ˛ = limn→∞ (n+1)n! · =
˛ ˛
an ˛ n→∞ ˛ ‹ nn
n
n n!
“ ”n “ ”n “ ”x
limn→∞ n+1n
= lim n→∞ 1 + 1
n
= limx→∞ 1 + x1 .
“ ”x
I limx→∞ 1 + x1 = limx→∞ e x ln(1+1/x) = e limx→∞ x ln(1+1/x) .
−1/x 2
limx→∞ x ln(1 + 1/x) = limx→∞ ln(1+1/x)
1/x
= (L0 Hop) limx→∞ (1+1/x)
−1/x 2
=
1
limx→∞ (1+1/x) = 1.
˛ ˛ “ ”x
˛a ˛
I Therefore limn→∞ ˛ n+1an ˛
= limx→∞ 1 + x1 = e 1 = e > 1 and the series
P∞ nn
n=1 n! diverges.

Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test

Example 4

P∞ (−1)n
Example 4 Test the following series for convergence n=1 n2
I We know already that this series converges absolutely and therefore it
converges. (we could also use the alternating series test to deduce this).
I Lets see what happens when we apply the ratio test here.
˛ ˛ ˛ 1‹(n+1)2 ˛ “ ”2
˛ an+1 ˛ n
I limn→∞ ˛ an ˛ = limn→∞ ˛ ‹ 2 ˛ = limn→∞ n+1 =
˛ ˛
1 n
“ ”2
1
limn→∞ 1+1/n = 1.
I Therefore the ratio test is inconclusive here.

Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test

The Root Test

Let ∞
P
Root Test n=1 an be a series (the terms may be positive or negative).
p P∞
I If limn→∞ n |an | = L < 1, then the series
n=1 an converges absolutely
(and hence is convergent).
p p P∞
I If limn→∞ n |an | = L > 1 or limn→∞ n |an | = ∞, then the series
n=1 an
is divergent.
p
I If limn→∞ n |an | = 1, then the Root test is inconclusive and we cannot
determine if the series converges or diverges using this test.
“ ”n
Example 5 Test the following series for convergence ∞ n−1 2n
P
n=1 (−1) n+1
r“ ”n
p 2n 2n 2
I limn→∞ n |an | = limn→∞ n = limn→∞ n+1 = limn→∞ 1+1/n =
n+1
2>1
P∞ “ ”n
n−1 2n
I Therefore by the n th root test, the series n=1 (−1) n+1
diverges.

Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test

Example 6

Root Test For ∞

P p
. L = limn→∞ n |an |.
n=1 anP
If L < 1, then the series ∞ an converges absolutely.
n=1 P
If L > 1 or ∞, then the series ∞ n=1 an is divergent.
If L = 1, then the Root test is inconclusive. “ ”n
Example 6 Test the following series for convergence ∞ n
P
n=1 2n+1
r“ ”n
p n n 1
I limn→∞ n |an | = limn→∞ n = limn→∞ 2n+1 = limn→∞ 2+1/n =
2n+1

1/2 < 1
P∞ “ ”n
n
I Therefore by the n th root test, the series n=1 2n+1
converges.

Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test

Example 7

Root Test For ∞

P p
. L = limn→∞ n |an |.
n=1 anP
If L < 1, then the series ∞ an converges absolutely.
n=1 P
If L > 1 or ∞, then the series ∞ n=1 an is divergent.
If L = 1, then the Root test is inconclusive. “ ”n
Example 7 Test the following series for convergence ∞ ln n
P
n=1 n
.
r“ ”
p n
ln n
I limn→∞ n |an | = limn→∞ n
n
= limn→∞ lnnn = limx→∞ lnxx =
1/x
(L0 Hop) limx→∞ 1
=0<1
P∞ “ ln n ”n
I Therefore by the n th root test, the series n=1 n
converges.

Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test

Rearranging sums

If we rearrange the terms in a finite sum, the sum remains the same. This is
not always the case for infinite sums (infinite series). It can be shown that:
P P
I If a series an is P
an absolutely convergent series with an = s, then any
rearrangement of an is convergent with sum s.
P
I It a series an is a conditionally convergent series, then for any real
P
number r , there is a rearrangement of an which has sum r .
(−1)n
Example The series ∞
P
I
n=1 2n is absolutely convergent with
P∞ (−1)n
n=1 2n = 3 and hence any rearrangement of the terms has sum 23 .
2

Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test

Rearranging sums
P
I It a series an is a conditionally convergent
P series, then for any real
number r , there is a rearrangement of an which has sum r .
P∞ (−1)n−1
I Example Alternating Harmonic series n=1 n
is conditionally
convergent, it can be shown that its sum is ln 2,
1 1 1 1 1 1 1 1 1
1− + − + − + − + − · · · + (−1)n + · · · = ln 2.
2 3 4 5 6 7 8 9 n
I Now we rearrange the terms taking the positive terms in blocks of one
followed by negative terms in blocks of 2
1 1 1 1 1 1 1 1 1
1−− + − − + − − + ··· =
2 4 3 6 8 5 10 12 7
“ 1” 1 “1 1” 1 “1 1” 1 “1 1”
1− − + − − + − − + − − ··· =
2 4 3 6 8 5 10 12 7 14
I
1“ 1 1 1 1 1 1 1 1 1 1
1 − + − + − + − + − · · · + (−1)n + . . . ) = ln 2.
2 2 3 4 5 6 7 8 9 n 2
I Obviously, we could continue in this way to get the series to sum to any
number of the form (ln 2)/2n .
Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test