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is
(a) increasing if f (x) > 0 for all x > m;
(b) decreasing if f (x) < 0 for all x > m;
(c) nondecreasing if f (x) ≥ 0 for all x > m;
(d) nonincreasing if f (x) ¤ 0 for all x > m.
Proof. Suppose that m ¤ a < b. Then by the Mean Value Theorem for
derivatives, there exists some c such that a < c < b and
f (b) ’ f (a)
= f (c),
b’a
f (b) = f (a) + f (c)(b ’ a).
The theorem follows from the above equation by considering the value of
f (c). In particular, for all natural numbers n ≥ m,
f (n + 1) = f (n) + f (c),
for some c such that n < c < n + 1.
Part (a). If f (c) > 0, then f (n + 1) > f (n) for all n ≥ m.
Part (b). If f (c) < 0, then f (n + 1) < f (n) for all n ≥ m.
Part (c). If f (c) ≥ 0, then f (n + 1) ≥ f (n) for all n ≥ m.
Part (d). If f (c) ¤ 0, then f (n + 1) ¤ f (n) for all n ¤ m.
This completes the proof of this theorem.
8.3. INFINITE SERIES 323

8.3 In¬nite Series
De¬nition 8.3.1 Let {tn }∞ be a given sequence. Let
n=1
n
s1 = t1 , s2 = t1 + t2 , s3 = t1 + t2 + t3 , · · · , sn = tk ,
k=1

for all natural number n. If the sequence {sn }∞ converges to a ¬nite number
n=1
L, then we write

L = t1 + t2 + t3 + · · · = tk .
k=1
n
We call tk an in¬nite series and write
k=1
∞ n
tk = lim tk = L.
n’∞
k=1 k=1

We say that L is the sum of the series and the series converges to L. If a
series does not converge to a ¬nite number, we say that it diverges. The
sequence {sn }∞ is called the sequence of the nth partial sums of the series.
n=1


Theorem 8.3.1 Suppose that a and r are real numbers and a = 0. Then
the geometric series

a
a + ar + ar2 + · · · = ark = ,
1’r
k=0

if |r| < 1. The geometric series diverges if |r| ≥ 1.
Proof. For each natural number n, let
sn = a + ar + · · · + arn’1 .
On multiplying both sides by r, we get
rsn = ar + ar2 + · · · + arn’1 + arn
sn ’ rsn = a ’ arn
(1 ’ r)sn = a(1 ’ rn )
a a
rn .

sn =
1’r 1’r
324 CHAPTER 8. INFINITE SERIES

If |r| < 1, then
a a a
lim rn =

lim sn = .
1 ’ r 1 ’ r n’∞ 1’r
n’∞

If |r| > 1, then lim rn is not ¬nite and so the sequence {sn }∞ of nth partial
n=1
n’∞
sums diverges.
If r = 1, then sn = na and lim na is not a ¬nite number.
n’∞
This completes the proof of the theorem.

Theorem 8.3.2 (Divergence Test) If the series tk converges, then lim tn =
n’∞
k=1
0. If lim tn = 0, then the series diverges.
n’∞

Proof. Suppose that the series converges to L. Then
n n’1
ak ’
lim an = lim ak
n’∞ n’∞
k=1 k=1
n n’1
ak ’ lim
= lim ak
n’∞ n’∞
k=1 k=1
=L’L
= 0.
The rest of the theorem follows from the preceding argument. This completes
the proof of this theorem.

Theorem 8.3.3 (The Integral Test) Let f be a function that is de¬ned,
continuous and decreasing on [1, ∞) such that f (x) > 0 for all x ≥ 1. Then
∞ ∞
f (n) and f (x)dx
1
n=1

either both converge or both diverge.
Proof. Suppose that f is decreasing and continuous on [1, ∞), and f (x) > 0
for all x ≥ 1. Then for all natural numbers n, we get,
n+1 n
n+1
f (k) ¤ f (x)dx ¤ f (k)
1
k=2 k=1
8.3. INFINITE SERIES 325

graph



It follows that,
∞ ∞

f (k) ¤ f (x)dx ¤ f (k).
1
k=2 k=1

Since f (1) is a ¬nite number, it follows that
∞ ∞
f (k) and f (x)dx
1
k=1

either both converge or both diverge. This completes the proof of the theo-
rem.

Theorem 8.3.4 Suppose that p > 0. Then the p-series

1
np
n=1

converges if p > 1 and diverges if 0 < p ¤ 1. In particular, the harmonic
series ∞ n diverges.
1
n=1

Proof. Suppose that p > 0. Then
∞ ∞
1
x’p dx
dx =
xp
1 1

x1’p
=
1’p 1
1
lim x1’p ’ 1 .
=
1’p x’∞

It follows that the integral converges if p > 1 and diverges if p < 1. If p = 1,
then ∞

1
dx = ln x = ∞.
x
1 1
Hence, the p-series converges if p > 1 and diverges if 0 < p ¤ 1. This
completes the proof of this theorem.
326 CHAPTER 8. INFINITE SERIES

Exercises 8.1
1. De¬ne the statement that the sequence {an }∞ converges to L.
n=1

2. Suppose the sequence {an }∞ converges to L and the sequences {bn }∞
n=1 n=1
converges to M . Then prove that
(a) {can }∞ converges to cL, where c is constant.
n=1
(b) {an + bn }∞ converges to L + M .
n=1
(c) {an ’ bn }∞ converges to L ’ M .
n=1
(d) {an bn }∞ converges to LM .
n=1

L
an
converges to , if M = 0.
(e)
bn M
n=1

3. Suppose that 0 < an ¤ an+1 < M for each natural number n. Then
prove that
(a) {an }∞ converges.
n=1
(b) {’an }∞ converges.
n=1

ak
(c) converges for each natural number k.
n n=1


xn
converges to 0 for every real number x.
4. Prove that
n! n=1

n!
5. Prove that converges to 0.
nn n=1

6. Prove that for each natural number n ≥ 2,
11 1 1 1
+ + · · · + < ln(n) < 1 + + · · · + .
(a)
n’1
23 n 2
1 1 1 n1 1 1
(b) p + p + · · · + p < 1 p dt < 1 + p + · · · + for each
(n ’ 1)p
2 3 n t 2
p > 0.

n ∞
1
1
(c) converges if and only if dt converges. De-
kp tp
1
k=1 n=1

n
1
termine the numbers p for which converges.
kp
n=1 n=1
8.4. SERIES WITH POSITIVE TERMS 327

n
rk converges if and only if |r| < 1.
7. Prove that
k=0 n=1


n
1
8. Prove that diverges.
k
k=1 n=1


n
1
9. Prove that diverges.
k ln k
k=2 n=2

10. Prove that for each natural number m ≥ 2,
m m+1
(a) (ln t)dt < ln(m!) < (ln t)dt
1 1
(b) m(ln(m) ’ 1) < ln(m!) < (m + 1)(ln(m + 1) ’ 1).
mm (m + 1)m+1
(c) m’1 < m! < .
em
e
(d) lim (m!)1/m = +∞.
m’+∞

(m!)1/m 1
(e) lim =
m e
m’+∞


11. Prove that {(’1)n }∞ does not converge.
n=1


sin(1/n)
12. Prove that converges to 1.
(1/n) n=1


sin n
13. Prove that converges to zero.
n n=1



8.4 Series with Positive Terms


k=1 bk
ak and
Theorem 8.4.1 (Algebraic Properties) Suppose that k=1
are convergent series and c > 0. Then
∞ ∞ ∞
(i) (ak + bk ) = ak + bk
k=1 k=1 k=1
328 CHAPTER 8. INFINITE SERIES

∞ ∞ ∞
(ak ’ bk ) = ak ’
(ii) bk
k=1 k=1 k=1

∞ ∞
(iii) c ak = c ak
k=1 k=1


(iv) If m is any natural number, then the series
∞ ∞
ck and ck
k=1 k=m


either both converge or both diverge.

Proof.
Part (i)
∞ n
(ak ± bk ) = lim (ak ± bk )
n’∞
k=1 k=1
n
n
± lim bk
= lim ak
n’∞
n’∞
k=1
k=1
∞ ∞
ak ±
= bk .
k=1 k=1


Part (ii) This part also follows from the preceding argument.

Part(iii) We see that
∞ n
c ak = lim c ak
n’∞
k=1 k=1
n
=c lim ak
n’∞
k=1

=c ak .
k=1
8.4. SERIES WITH POSITIVE TERMS 329

Part (iv) We observe that
∞ ∞
m’1
ak = ak + ak .
k=1 k=1 k=1



Therefore,
∞ n
ak = lim ak
n’∞
k=1 k=1
m’1 n
= ak + lim ak .
n’∞
k=1 k=m

It follows that the series
∞ ∞
ak and ak
k=1 k=m

either both converge or both diverge. This completes the proof of this theo-
rem.

Theorem 8.4.2 (Comparison Test) Suppose that 0 < an ¤ bn for all natural
numbers n ≥ 1.

(a) If there exists some M such that n ak ¤ M , for all natural numbers
k=1
n, then ∞ ak converges. If there exists no such M , then the series
k=1
diverges.


ak converges.
k=1 bk converges, then
(b) If k=1



k=1 bk diverges.
ak diverges, then
(c) If k=1

(d) If cn > 0 for all natural numbers n, and
cn
= L, 0 < L < ∞,
lim
an
n’∞



k=1 ck either both converge or both diverge.
ak and
then the series k=1
330 CHAPTER 8. INFINITE SERIES

n n
bk , 0 < an ¤ bn for all natural numbers
Proof. Let An = ak , Bn =
k=1 k=1
{An }∞
and {Bn }∞ are strictly increasing sequence. Let
n. The sequences n=1 n=1
A represent the least upper bound of {An }∞ and let B represent the least
n=1

upper bound of {Bn }n=1

Part (a) If An ¤ M for all natural numbers, then {An }∞ is a bounded
n=1
and strictly increasing sequence. Then A is a ¬nite number and {An }∞
n=1
converges to A and

A= ak .
k=1



∞ ∞
bk = B and An ¤ Bn ¤ B for all
Part (b) If bk converges, then
k=1 k=1

natural numbers n. By Part (a), ak converges to A.
k=1


ak diverges, then the sequence {An }∞ diverges. Since {An }∞
Part (c) If n=1 n=1
k=1
is strictly increasing and divergent, for every M there exists some m such
that
M < An ¤ Bn

for all natural numbers n ≥ m. It follows that {Bn }∞ diverges.
n=1

L
Part (d) Suppose that 0 < an and 0 < cn , 0 < L < ∞, and
=
2

cn
lim = L.
an
n’∞


Then there exists some natural number m such that

cn 1
’L <
an 2
8.4. SERIES WITH POSITIVE TERMS 331

for all natural numbers n ≥ m. Hence, for all n ≥ m, we have

L cn L L cn 3
’ ’L< ,
< < <L
2 an 2 2 an 2
3
L
an ¤ cn ¤ L an .
2 2
n n
3
L
ak ¤ mck ¤ L ak
2 2
k=m k=m k=m

∞ ∞
n m n
L
If diverges, then diverges and, hence
ak ak ck
2
k=1 k=m k=m
n=1 n=m

n
and both diverge.
ck
k=1 k=1

∞ n
n
3
converges and, hence,
If converges, then L ak
ak
2 k=m
k=1 n=m
k=1
∞ ∞
n
and both converge.
ck ck
k=m k=1
n=m n=1
This completes the Proof of Theorem 8.4.2.

Theorem 8.4.3 (Ratio Test) Suppose that 0 < an for every natural number
n and
an+1
lim = r.
an
n’∞


ak
Then the series k=1


(a) converges if r < 1;

(b) diverges if r > 1;

(c) may converge or diverge if r = 1; the test fails.

Proof. Suppose that 0 < an for every natural number n and
an+1
lim = r.
an
n’∞
332 CHAPTER 8. INFINITE SERIES

Let > 0 be given. Then there exists some natural number M such that

an+1 an+1
’r < , ’ +r < <r+
an an
(r ’ )an < an+1 < (r + )an (1)

for all natural numbers n ≥ M .
Part (a) Suppose that 0 ¤ r < 1 and = (1 ’ r)/2. Then for each natural
number k, we have
k
1+r
k
am+k < (r + ) am = am . . . (2)
2

Hence, by (2), we get
∞ ∞
m’1
an = an + am+k
n=1 n=1 k=0

m’1 k
1+r
am
< an +
2
n=1 k=0
m’1
am
= an +
1 ’ 1+r
2
n=1
m’1
2am
= an +
1’r
n=1
< ∞.

It follows that the series an converges.
n=1

= (r ’ 1)/2. Then by (1) we get
Part (b) Suppose that 1 < r,

3r ’ 1
an < an < an+1
2
for all n ≥ m. It follows that

0 < am ¤ lim am+k = lim an .
n’∞
k’∞
8.4. SERIES WITH POSITIVE TERMS 333


By the Divergence test, the series an diverges.
n=1


1
1
and ,
Part (c) For both series
n2
n n=1
n=1
an+1
lim = 1.
an
n’∞

∞ ∞
1 1
But, by the p-series test, diverges and converges. Thus, the
n2
n
n=1 n=1
ratio test fails to test the convergence or divergence of these series when
r = 1.
This completes the proof of Theorem 8.4.3.

Theorem 8.4.4 (Root Test) Suppose that 0 < an for each natural number
n and
lim (an )1/n = r.
n’∞

ak
Then the series k=1

(a) converges if r < 1;
(b) diverges if r > 1;
(c) may converge or diverge if r = 1; the test fails.
Proof. Suppose that 0 < an for each natural number n and
lim (an )1/n = r.
n’∞

Let > 0 be given. Then there exists some natural number m such that
(an )1/n ’ r <
r ’ < (an )1/n < r + . . . (3)
for all natural numbers n ≥ m.
1+r
. Then, by (3), for each natural number
Part (a) Suppose r < 1 and =
2
n ≥ m, we have
n
1’r
1+r
(an )1/n < and an < .
2 2
334 CHAPTER 8. INFINITE SERIES

it follows that
∞ ∞
m’1
ak = an + an
n=1 n=1 n=m

m’1 n
1+r
< an +
2
n=1 n=m
m’1 m
1
1+r
= an + 1+r
1’
2 2
n=1
m’1 m
1+r 2
= an +
1’r
2
n=1
< ∞.

Therefore, ak converges.
n=1

= (r ’ 1)/2. Then, by (3), for each natural
Part (b) Suppose r > 1 and
number n ≥ m, we have
1+r
= r + < (an )1/n
1<
2
n
1+r
1< < an .
2

It follows that lim an = 0 and, by the Divergence test, the series an
n’∞
n=1
diverges.


1
1
and we have r = 1, where
Part (c) For each of the series
n2
n n=1
n=1

r = lim (an )1/n .
n’∞


1
1
diverges and the series converges by the p-series
But the series
n2
n n=1
n=1
test. Therefore, the test fails to determine the convergence or divergence for
these series when r = 1. This completes the proof of Theorem 8.4.4.
8.4. SERIES WITH POSITIVE TERMS 335

Exercises 8.2

1. De¬ne what is meant by ak .
k=1

2. De¬ne what is meant by the sequence of nth partial sums of the series

ak .
k=1

a
ark converges to if |r| < 1.
3. Suppose that a = 0. Prove that
1’r
k=0

3
1
4. Prove that the series converges to .
k(k + 2) 4
k=1

1
1
5. Prove that converges to if p > 1 and diverges otherwise.
kp p’1
k=1


n n
6. Prove that is an increasing sequence and the series ln
n+1 n+1
n=1 n=1
diverges.

1
(’1)k xk converges to if |x| < 1.
7. Prove that
1+x
k=0

1
x2k converges to if |x| < 1.
8. Prove that
1 ’ x2
k=0

1
(’1)k x2k converges to if |x| < 1.
9. Prove that
1 + x2
k=0

10. Prove that if ak converges, then lim ak = 0. Is the converse true?
k’∞
k=0
Explain your answer.
∞ ∞
11. Suppose that if ak converges to L and bk converges to M . Prove
k=0 k=0
that
336 CHAPTER 8. INFINITE SERIES


(a) (c ak ) converges to cL for each constant c.
k=0

(b) (ak + bk ) converges to L + M .
k=0

(ak ’ bk ) converges to L ’ M .
(c)
k=0

(d) ak bk may or may not converge to LM .
k=0

∞ ∞
1
1
converges if and only if dt converges. Deter-
12. Prove that
kp tp
1
k=1
mine the values of p for which the series converges.

13. Suppose that f (x) is continuous and decreasing on the interval [a, +∞).

Let ak = f (k) for each natural number k. Then the series ak con-
k=1

verges if and only if f (x)dx converges.
a
n
14. Suppose that 0 ¤ ak ¤ ak+1 for each natural number k, and sn = ak .
k=1

Prove that if sn ¤ M for some M and all natural numbers n, then ak
k=1
converges.

15. Suppose that 0 ¤ ak ¤ bk for each natural number k. Prove that
∞ ∞
(a) if bk converges, then ak converges.
k=1 k=1
∞ ∞
(b) if ak diverges, then bk diverges.
k=1 k=1

(c) if lim ak = 0, then ak diverges.
k’∞
k=1
8.4. SERIES WITH POSITIVE TERMS 337


(d) if lim ak = 0, then ak may or may not converge.
k’∞
k=1


16. Suppose that 0 < ak for each natural number k. Prove that if lim (ak+1 /ak ) <
k’∞

1, then ak converges.
k=1

17. Suppose that 0 < ak for each natural number k. Prove that if lim (ak+1 /ak ) >
k’∞

1, then ak diverges.
k=1

18. Suppose that 0 < ak for each natural number k. Prove that if lim (ak+1 /ak ) =
k’∞

1, then ak may or may not converge.
k=1

19. Suppose that 0 < ak and 0 < bk for each natural number k. Prove that
∞ ∞
if 0 < lim (ak /bk ) < ∞, then ak converges if and only if bk
k’∞
k=1 k=1
converges.
20. Suppose that 0 < ak for each natural number k. Prove that if lim (ak )1/k <
k’∞

1, then ak converges.
k=1

21. Suppose that 0 < ak for each natural number k. Prove that if lim (ak )1/k >
k’∞

1, then ak diverges.
k=1

22. Suppose that 0 < ak for each natural number k. Prove that if lim (ak )1/k =
k’∞

1, then ak may or may not converge.
k=1
∞ ∞
|ak | converges. Sup-
23. A series ak is said to converge absolutely if
k=1 k=1
pose that lim |ak+1 /ak | = p. Prove that
k’∞
338 CHAPTER 8. INFINITE SERIES


(a) ak converges absolutely if p < 1.
k=1

(b) ak does not converge absolutely if p > 1.
k=1

(c) ak may or may not converge absolutely if p = 1.
k=1


∞ ∞
|ak | converges. Sup-
24. A series ak is said to converge absolutely if
k=1 k=1
1/k
pose that lim (|ak |) = p. Prove that
k’∞


(a) ak converges absolutely if p < 1.
k=1

(b) ak does not converge absolutely if p > 1.
k=1

(c) ak may or may not converge absolutely if p = 1.
k=1



25. Prove that if ak converges absolutely, then it converges. Is the
k=1
converse true? Justify your answer.

ak
= p.
26. Suppose that ak = 0, bk = 0 for any natural number k and lim
bk
k’∞

Prove that if 0 < p < 1, then the series ak converges absolutely if
k=1

and only if bk converges absolutely.
k=1

∞ ∞
27. A series ak is said to converge conditionally if ak converges but
k=1 k=1
8.4. SERIES WITH POSITIVE TERMS 339

∞ ∞
(’1)n+1
|ak | diverges. Determine whether the series converges
n
n=1
k=1
conditionally or absolutely.

28. Suppose that 0 < ak and |ak+1 | < |ak | for every natural number k. Prove
∞ ∞
k+1
(’1)k ak are
that if lim ak = 0, then the series (’1) ak and
k’+∞
k=1 k=1
both convergent. Furthermore, show that if s denotes the sum of the
series, then s is between the nth partial sum sn and the (n + 1)st partial
sum sn+1 for each natural number n.

n
(’1)n converges absolutely or condi-
29. Determine whether the series
3n
n=1
tionally.

(2n)!
(’1)n
30. Determine whether the series converges absolutely or con-
n10
n=1
ditionally.

In problems 31“62, test the given series for convergence, conditional conver-
gence or absolute convergence.
∞ ∞ n
n! n+1 5
(’1)n n
31. 32. (’1)
5 n!
n=1 n=1

∞ ∞
n n
4 4
n n+1 2
33. 34.
(’1) n (’1) n
5 5
n=1 n=1

∞ ∞
(’1)n (’1)n+1
35. 36.
n3/2 n1/2
n=1 n=1

∞ ∞
(’1)n (’1)n+1
37. ,0 < p < 1 38. ,1 < p
np np
n=1 n=1

∞ ∞
(n + 1)2
(n + 1)
(’1)n 2 n+1
39. 40. (’1)
3n
n +2
n=1 n=1
340 CHAPTER 8. INFINITE SERIES


3n
∞ ∞
(n + 2)2
(’1)n+1 (’1)n’1 2
41. 42.
(n + 1)3 n2
n=1 n=1

∞ ∞
(’1)n (4/3)n (’4)n
43. 44.
n4 (n!)n
n=1 n=1

∞ ∞
n (n + 1)!
n
(’1)n
45. 46.
(’3)
1 · 3 · 5 · · · (2n + 1)
(2n)!
n=1 n=1

∞ ∞
2n
(n ’ 1)
n (n!) 2
(’1)n+1
47. 48.
(’1)
n3/2
(2n)!
n=1 n=1

∞ ∞
4n 2 · 4 · · · (2n + 2)
n 2
(’1)n
49. 50.
(’1) (n!)
1 · 4 · 7 · · · (3n + 1)
(2n)!
n=1 n=1

∞ ∞
n+1
n’1 5 (n + 1)
(’1)n+1
51. 52.
(’1)
24n (n + 3)
n=1 n=1

∞ ∞
n+1 (n + 2) (n + 2)
(’1)n
53. 54.
(’1)
n5/4 n7/4
n=1 n=1

∞ ∞
2
(’1)n
+ 2n ’ 1)
n (3n
55. 56.
(’1)
2n3 n(ln n)
n=1 n=2

∞ ∞
n (ln n) (ln n)
(’1)n+1
57. 58.
(’1)
n2
n
n=2 n=1

∞ ∞ p
n! nn
n
59. (’1) p , 0 < p < 1 60. ,0 < p < 1
(’1)
n n!
n=1 n=1

∞ ∞ p
n! nn
n
61. (’1) p , 1 < p 62. ,1 < p
(’1)
n n!
n=1 n=1
8.5. ALTERNATING SERIES 341


63. Suppose that 0 < ak for each natural number k and ak converges.
k=1

ap converges for every p > 1.
Prove that k
k=1


64. Suppose that 0 < ak for each natural number k and ak diverges.
k=1

ap , for 0 < p < 1.
Prove that k
k=1


65. Suppose that 0 < r < 1 and |ak+1 /ak | < r for all k ≥ N . Prove that

ak converges absolutely.
k=1


an
n
converges absolutely if 0 < a < b.
66. Prove that (’1)
3 + bn
k=1



8.5 Alternating Series
De¬nition 8.5.1 Suppose that for each natural number n, bn is positive or
negative. Then the series ∞ bk is said to converge
k=1


|bk | converges;
(a) absolutely if the series k=1



|bk | diverges.
k=1 bk converges but
(b) conditionally if the series k=1



Theorem 8.5.1 If a series converges absolutely, then it converges.

|bk | converges. For each natural number k, let
Proof. Suppose that
k=1
ak = bk + |bk | and ck = 2|bk |. Then 0 ¤ ak ¤ ck for each k. Since
∞ ∞ ∞
2|bk | = 2 |bk |,
ck =
k=1 k=1 k=1
342 CHAPTER 8. INFINITE SERIES

∞ ∞
the series ck converges. by the comparison test ak also converges. It
k=1 k=0
follows that
∞ ∞
(ak ’ |bk |)
bk =
k=1 k=1
∞ ∞
ak ’ |bk |
=
k=1 k=1


and the series bk converges. This completes the proof of the theorem.
k=1


De¬nition 8.5.2 Suppose that for each natural number n, an > 0. Then an
alternating series is a series that has one of the following two forms:
n
n+1
(’1)k+1 ak
(a) a1 ’ a2 + a3 ’ · · · + (’1) an + · · · =
k=1


n
(’1)k ak .
(b) ’a1 + a2 ’ a3 + · · · + (’1) an + · · · =
k=1



Theorem 8.5.2 Suppose that 0 < an+1 < an for all natural numbers m, and
lim an = 0. Then
n’∞

∞ ∞
n
(’1)n+1 an both converge.
(a) (’1) an and
n=1 n=1

∞ n
k+1
(’1)k+1 an < an+1 , for all n;
an ’
(b) (’1)
k=1 k=1

∞ ∞
(’1)k ak ’ (’1)k ak < an+1 , or all n.
(c)
k=1 k=1

Proof.
8.5. ALTERNATING SERIES 343

n
(’1)k+1 ak . Then,
Part (a) For each natural number n, let sn =
k=1

2n+2 2n
k+1
(’1)k+1 ak
s2n+2 ’ s2n = ak ’
(’1)
k=1 k=1
2n+3
a2n+2 + (’1)2n+2 a2n+1
= (’1)
= a2n+1 ’ a2n+2 > 0.

Therefore, s2n+2 > s2n and {s2n }∞ is an increasing sequence. Similarly,
n=1

s2n+3 ’ s2n+1 = (’1)2n+4 a2n+3 ’ (’1)2n+2 a2n+1 = a2n+3 ’ a2n+1 < 0.

Therefore, s2n+3 < s2n+1 and {s2n+1 }∞ is a decreasing sequence. Further-
n=0
more,

s2n = a1 ’ a2 + a3 ’ a4 + · · · + (’1)2n+1 a2n
= a1 ’ (a2 ’ a3 ) ’ (a4 ’ a5 ) ’ · · · ’ (a2n’2 ’ a2n’1 ) ’ a2n < a1 .

Thus, {s2n }∞ is an increasing sequence which is bounded above by a1 .
n=1
Therefore, {s2n }∞ converges to some number s ¤ a1 . Then
n=1

lim s2n+1 = lim s2n + lim a2n+1
n’∞ n’∞ n’∞
= lim s2n
n’∞
= s.

It follows that
lim sn = s
n’∞
∞ ∞
(’1)n+1 ak converges to s and the series (’1)n ak con-
and the series
n=1 n=1
verges to ’s.
Part (b) In the proof of Part (a) we showed that

s2n < s < s2n+1 < s2n’1 . . . (1)

for each natural number n. It follows that

0 < s ’ s2n < s2n+1 ’ s2n = a2n+1
344 CHAPTER 8. INFINITE SERIES

and
∞ 2n
(’1)k+1 ak ’ (’1)k+1 ak < a2n+1 .
k=1 k=1

Similarly,

s2n ’ s2n’1 < s ’ s2n’1
s2n’1 ’ s2n > s2n’1 ’ s
s ’ s2n’1 < s2n’1 ’ s2n = a2n
∞ 2n’1
k+1
(’1)k+1 ak < a2n .
ak ’
(’1)
k=1 k=1

It follows that for all natural numbers n,
∞ n
k+1
(’1)k+1 ak < an+1 .
ak ’
(’1)
k=1 k=1

∞ n
(’1)k ak ’ (’1)k ak
Part (c)
k=1 k=1

∞ n
(’1)k+1 ak ’ (’1)k+1 ak
= (’1)
k=1 k=1
∞ n
k+1
(’1)k+1 ak < a2n+1 .
ak ’
= (’1)
k=1 k=1

This concludes the proof of this theorem.

ak . Let
Theorem 8.5.3 Consider a series k=1

an+1
= L , lim |an |1/n = M
lim
an
n’∞ n’∞



ak converges absolutely.
(a) If L < 1, then the series k=1

ak does not converge absolutely.
(b) If L > 1, then the series k=1

ak converges absolutely.
(c) If M < 1, then the series k=1
8.5. ALTERNATING SERIES 345


ak does not converge absolutely.
(d) If M > 1, then the series k=1

ak may or may not converge
(e) If L = 1 or M = 1, then the series k=1
absolutely.

Proof. Suppose that for a series ak ,
k=1

an+1
lim |an |1/n = M.
lim = L and
an

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