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

8.3 Inп¬Ѓnite Series
Deп¬Ѓnition 8.3.1 Let {tn }в€ћ be a given sequence. Let
n=1
n
s1 = t1 , s2 = t1 + t2 , s3 = t1 + t2 + t3 , В· В· В· , sn = tk ,
k=1

for all natural number n. If the sequence {sn }в€ћ converges to a п¬Ѓnite number
n=1
L, then we write
в€ћ
L = t1 + t2 + t3 + В· В· В· = tk .
k=1
n
We call tk an inп¬Ѓnite series and write
k=1
в€ћ n
tk = lim tk = L.
nв†’в€ћ
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 п¬Ѓnite number, we say that it diverges. The
sequence {sn }в€ћ 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
в€ћ
a
a + ar + ar2 + В· В· В· = ark = ,
1в€’r
k=0

if |r| < 1. The geometric series diverges if |r| в‰Ґ 1.
Proof. For each natural number n, let
sn = a + ar + В· В· В· + arnв€’1 .
On multiplying both sides by r, we get
rsn = ar + ar2 + В· В· В· + arnв€’1 + arn
sn в€’ rsn = a в€’ arn
(1 в€’ r)sn = a(1 в€’ rn )
a a
rn .
в€’
sn =
1в€’r 1в€’r
324 CHAPTER 8. INFINITE SERIES

If |r| < 1, then
a a a
lim rn =
в€’
lim sn = .
1 в€’ r 1 в€’ r nв†’в€ћ 1в€’r
nв†’в€ћ

If |r| > 1, then lim rn is not п¬Ѓnite and so the sequence {sn }в€ћ of nth partial
n=1
nв†’в€ћ
sums diverges.
If r = 1, then sn = na and lim na is not a п¬Ѓnite number.
nв†’в€ћ
This completes the proof of the theorem.
в€ћ
Theorem 8.3.2 (Divergence Test) If the series tk converges, then lim tn =
nв†’в€ћ
k=1
0. If lim tn = 0, then the series diverges.
nв†’в€ћ

Proof. Suppose that the series converges to L. Then
n nв€’1
ak в€’
lim an = lim ak
nв†’в€ћ nв†’в€ћ
k=1 k=1
n nв€’1
ak в€’ lim
= lim ak
nв†’в€ћ nв†’в€ћ
k=1 k=1
=Lв€’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 deп¬Ѓned,
continuous and decreasing on [1, в€ћ) such that f (x) > 0 for all x в‰Ґ 1. Then
в€ћ в€ћ
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, в€ћ), and f (x) > 0
for all x в‰Ґ 1. Then for all natural numbers n, we get,
n+1 n
n+1
f (k) в‰¤ f (x)dx в‰¤ f (k)
1
k=2 k=1
8.3. INFINITE SERIES 325

graph

It follows that,
в€ћ в€ћ
в€ћ
f (k) в‰¤ f (x)dx в‰¤ f (k).
1
k=2 k=1

Since f (1) is a п¬Ѓnite number, it follows that
в€ћ в€ћ
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
в€ћ
1
np
n=1

converges if p > 1 and diverges if 0 < p в‰¤ 1. In particular, the harmonic
series в€ћ n diverges.
1
n=1

Proof. Suppose that p > 0. Then
в€ћ в€ћ
1
xв€’p dx
dx =
xp
1 1
в€ћ
x1в€’p
=
1в€’p 1
1
lim x1в€’p в€’ 1 .
=
1в€’p xв†’в€ћ

It follows that the integral converges if p > 1 and diverges if p < 1. If p = 1,
then в€ћ
в€ћ
1
dx = ln x = в€ћ.
x
1 1
Hence, the p-series converges if p > 1 and diverges if 0 < p в‰¤ 1. This
completes the proof of this theorem.
326 CHAPTER 8. INFINITE SERIES

Exercises 8.1
1. Deп¬Ѓne the statement that the sequence {an }в€ћ converges to L.
n=1

2. Suppose the sequence {an }в€ћ converges to L and the sequences {bn }в€ћ
n=1 n=1
converges to M . Then prove that
(a) {can }в€ћ converges to cL, where c is constant.
n=1
(b) {an + bn }в€ћ converges to L + M .
n=1
(c) {an в€’ bn }в€ћ converges to L в€’ M .
n=1
(d) {an bn }в€ћ converges to LM .
n=1
в€ћ
L
an
converges to , if M = 0.
(e)
bn M
n=1

3. Suppose that 0 < an в‰¤ an+1 < M for each natural number n. Then
prove that
(a) {an }в€ћ converges.
n=1
(b) {в€’an }в€ћ converges.
n=1
в€ћ
ak
(c) converges for each natural number k.
n n=1

в€ћ
xn
converges to 0 for every real number x.
4. Prove that
n! n=1
в€ћ
n!
5. Prove that converges to 0.
nn n=1

6. Prove that for each natural number n в‰Ґ 2,
11 1 1 1
+ + В· В· В· + < ln(n) < 1 + + В· В· В· + .
(a)
nв€’1
23 n 2
1 1 1 n1 1 1
(b) p + p + В· В· В· + p < 1 p dt < 1 + p + В· В· В· + for each
(n в€’ 1)p
2 3 n t 2
p > 0.
в€ћ
n в€ћ
1
1
(c) converges if and only if dt converges. De-
kp tp
1
k=1 n=1
в€ћ
n
1
termine the numbers p for which converges.
kp
n=1 n=1
8.4. SERIES WITH POSITIVE TERMS 327
в€ћ
n
rk converges if and only if |r| < 1.
7. Prove that
k=0 n=1

в€ћ
n
1
8. Prove that diverges.
k
k=1 n=1

в€ћ
n
1
9. Prove that diverges.
k ln k
k=2 n=2

10. Prove that for each natural number m в‰Ґ 2,
m m+1
(a) (ln t)dt < ln(m!) < (ln t)dt
1 1
(b) m(ln(m) в€’ 1) < ln(m!) < (m + 1)(ln(m + 1) в€’ 1).
mm (m + 1)m+1
(c) mв€’1 < m! < .
em
e
(d) lim (m!)1/m = +в€ћ.
mв†’+в€ћ

(m!)1/m 1
(e) lim =
m e
mв†’+в€ћ

11. Prove that {(в€’1)n }в€ћ does not converge.
n=1

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

в€ћ
sin n
13. Prove that converges to zero.
n n=1

8.4 Series with Positive Terms
в€ћ
в€ћ
k=1 bk
ak and
Theorem 8.4.1 (Algebraic Properties) Suppose that k=1
are convergent series and c > 0. Then
в€ћ в€ћ в€ћ
(i) (ak + bk ) = ak + bk
k=1 k=1 k=1
328 CHAPTER 8. INFINITE SERIES

в€ћ в€ћ в€ћ
(ak в€’ bk ) = ak в€’
(ii) bk
k=1 k=1 k=1

в€ћ в€ћ
(iii) c ak = c ak
k=1 k=1

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

either both converge or both diverge.

Proof.
Part (i)
в€ћ n
(ak В± bk ) = lim (ak В± bk )
nв†’в€ћ
k=1 k=1
n
n
В± lim bk
= lim ak
nв†’в€ћ
nв†’в€ћ
k=1
k=1
в€ћ в€ћ
ak В±
= bk .
k=1 k=1

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

Part(iii) We see that
в€ћ n
c ak = lim c ak
nв†’в€ћ
k=1 k=1
n
=c lim ak
nв†’в€ћ
k=1
в€ћ
=c ak .
k=1
8.4. SERIES WITH POSITIVE TERMS 329

Part (iv) We observe that
в€ћ в€ћ
mв€’1
ak = ak + ak .
k=1 k=1 k=1

Therefore,
в€ћ n
ak = lim ak
nв†’в€ћ
k=1 k=1
mв€’1 n
= ak + lim ak .
nв†’в€ћ
k=1 k=m

It follows that the series
в€ћ в€ћ
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 в‰¤ bn for all natural
numbers n в‰Ґ 1.

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

в€ћ
в€ћ
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 < в€ћ,
lim
an
nв†’в€ћ

в€ћ
в€ћ
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 в‰¤ bn for all natural numbers
Proof. Let An = ak , Bn =
k=1 k=1
{An }в€ћ
and {Bn }в€ћ are strictly increasing sequence. Let
n. The sequences n=1 n=1
A represent the least upper bound of {An }в€ћ and let B represent the least
n=1
в€ћ
upper bound of {Bn }n=1

Part (a) If An в‰¤ M for all natural numbers, then {An }в€ћ is a bounded
n=1
and strictly increasing sequence. Then A is a п¬Ѓnite number and {An }в€ћ
n=1
converges to A and
в€ћ
A= ak .
k=1

в€ћ в€ћ
bk = B and An в‰¤ Bn в‰¤ B for all
Part (b) If bk converges, then
k=1 k=1
в€ћ
natural numbers n. By Part (a), ak converges to A.
k=1

в€ћ
ak diverges, then the sequence {An }в€ћ diverges. Since {An }в€ћ
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 в‰¤ Bn

for all natural numbers n в‰Ґ m. It follows that {Bn }в€ћ diverges.
n=1

L
Part (d) Suppose that 0 < an and 0 < cn , 0 < L < в€ћ, and
=
2

cn
lim = L.
an
nв†’в€ћ

Then there exists some natural number m such that

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

for all natural numbers n в‰Ґ m. Hence, for all n в‰Ґ m, we have

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

в€ћ в€ћ
n m n
L
If diverges, then diverges and, hence
ak ak ck
2
k=1 k=m k=m
n=1 n=m
в€ћ
n
and both diverge.
ck
k=1 k=1
в€ћ
в€ћ n
n
3
converges and, hence,
If converges, then L ak
ak
2 k=m
k=1 n=m
k=1
в€ћ в€ћ
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
nв†’в€ћ

в€ћ
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
nв†’в€ћ
332 CHAPTER 8. INFINITE SERIES

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

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

for all natural numbers n в‰Ґ M .
Part (a) Suppose that 0 в‰¤ r < 1 and = (1 в€’ 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
в€ћ в€ћ
mв€’1
an = an + am+k
n=1 n=1 k=0
в€ћ
mв€’1 k
1+r
am
< an +
2
n=1 k=0
mв€’1
am
= an +
1 в€’ 1+r
2
n=1
mв€’1
2am
= an +
1в€’r
n=1
< в€ћ.
в€ћ
It follows that the series an converges.
n=1

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

3r в€’ 1
an < an < an+1
2
for all n в‰Ґ m. It follows that

0 < am в‰¤ lim am+k = lim an .
nв†’в€ћ
kв†’в€ћ
8.4. SERIES WITH POSITIVE TERMS 333

в€ћ
By the Divergence test, the series an diverges.
n=1
в€ћ
в€ћ
1
1
and ,
Part (c) For both series
n2
n n=1
n=1
an+1
lim = 1.
an
nв†’в€ћ

в€ћ в€ћ
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.
nв†’в€ћ
в€ћ
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.
nв†’в€ћ

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

it follows that
в€ћ в€ћ
mв€’1
ak = an + an
n=1 n=1 n=m
в€ћ
mв€’1 n
1+r
< an +
2
n=1 n=m
mв€’1 m
1
1+r
= an + 1+r
1в€’
2 2
n=1
mв€’1 m
1+r 2
= an +
1в€’r
2
n=1
< в€ћ.
в€ћ
Therefore, ak converges.
n=1

= (r в€’ 1)/2. Then, by (3), for each natural
Part (b) Suppose r > 1 and
number n в‰Ґ m, we have
1+r
= r + < (an )1/n
1<
2
n
1+r
1< < an .
2
в€ћ
It follows that lim an = 0 and, by the Divergence test, the series an
nв†’в€ћ
n=1
diverges.
в€ћ
в€ћ
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 .
nв†’в€ћ
в€ћ
в€ћ
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
в€ћ
1. Deп¬Ѓne what is meant by ak .
k=1

2. Deп¬Ѓne what is meant by the sequence of nth partial sums of the series
в€ћ
ak .
k=1
в€ћ
a
ark converges to if |r| < 1.
3. Suppose that a = 0. Prove that
1в€’r
k=0
в€ћ
3
1
4. Prove that the series converges to .
k(k + 2) 4
k=1
в€ћ
1
1
5. Prove that converges to if p > 1 and diverges otherwise.
kp pв€’1
k=1
в€ћ
в€ћ
n n
6. Prove that is an increasing sequence and the series ln
n+1 n+1
n=1 n=1
diverges.
в€ћ
1
(в€’1)k xk converges to if |x| < 1.
7. Prove that
1+x
k=0
в€ћ
1
x2k converges to if |x| < 1.
8. Prove that
1 в€’ x2
k=0
в€ћ
1
(в€’1)k x2k converges to if |x| < 1.
9. Prove that
1 + x2
k=0
в€ћ
10. Prove that if ak converges, then lim ak = 0. Is the converse true?
kв†’в€ћ
k=0
в€ћ в€ћ
11. Suppose that if ak converges to L and bk converges to M . Prove
k=0 k=0
that
336 CHAPTER 8. INFINITE SERIES

в€ћ
(a) (c ak ) converges to cL for each constant c.
k=0
в€ћ
(b) (ak + bk ) converges to L + M .
k=0
в€ћ
(ak в€’ bk ) converges to L в€’ M .
(c)
k=0
в€ћ
(d) ak bk may or may not converge to LM .
k=0

в€ћ в€ћ
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, +в€ћ).
в€ћ
Let ak = f (k) for each natural number k. Then the series ak con-
k=1
в€ћ
verges if and only if f (x)dx converges.
a
n
14. Suppose that 0 в‰¤ ak в‰¤ ak+1 for each natural number k, and sn = ak .
k=1
в€ћ
Prove that if sn в‰¤ M for some M and all natural numbers n, then ak
k=1
converges.

15. Suppose that 0 в‰¤ ak в‰¤ bk for each natural number k. Prove that
в€ћ в€ћ
(a) if bk converges, then ak converges.
k=1 k=1
в€ћ в€ћ
(b) if ak diverges, then bk diverges.
k=1 k=1
в€ћ
(c) if lim ak = 0, then ak diverges.
kв†’в€ћ
k=1
8.4. SERIES WITH POSITIVE TERMS 337

в€ћ
(d) if lim ak = 0, then ak may or may not converge.
kв†’в€ћ
k=1

16. Suppose that 0 < ak for each natural number k. Prove that if lim (ak+1 /ak ) <
kв†’в€ћ
в€ћ
1, then ak converges.
k=1

17. Suppose that 0 < ak for each natural number k. Prove that if lim (ak+1 /ak ) >
kв†’в€ћ
в€ћ
1, then ak diverges.
k=1

18. Suppose that 0 < ak for each natural number k. Prove that if lim (ak+1 /ak ) =
kв†’в€ћ
в€ћ
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
в€ћ в€ћ
if 0 < lim (ak /bk ) < в€ћ, then ak converges if and only if bk
kв†’в€ћ
k=1 k=1
converges.
20. Suppose that 0 < ak for each natural number k. Prove that if lim (ak )1/k <
kв†’в€ћ
в€ћ
1, then ak converges.
k=1

21. Suppose that 0 < ak for each natural number k. Prove that if lim (ak )1/k >
kв†’в€ћ
в€ћ
1, then ak diverges.
k=1

22. Suppose that 0 < ak for each natural number k. Prove that if lim (ak )1/k =
kв†’в€ћ
в€ћ
1, then ak may or may not converge.
k=1
в€ћ в€ћ
|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
kв†’в€ћ
338 CHAPTER 8. INFINITE SERIES

в€ћ
(a) ak converges absolutely if p < 1.
k=1
в€ћ
(b) ak does not converge absolutely if p > 1.
k=1
в€ћ
(c) ak may or may not converge absolutely if p = 1.
k=1

в€ћ в€ћ
|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
kв†’в€ћ

в€ћ
(a) ak converges absolutely if p < 1.
k=1
в€ћ
(b) ak does not converge absolutely if p > 1.
k=1
в€ћ
(c) ak may or may not converge absolutely if p = 1.
k=1

в€ћ
25. Prove that if ak converges absolutely, then it converges. Is the
k=1

ak
= p.
26. Suppose that ak = 0, bk = 0 for any natural number k and lim
bk
kв†’в€ћ
в€ћ
Prove that if 0 < p < 1, then the series ak converges absolutely if
k=1
в€ћ
and only if bk converges absolutely.
k=1

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

в€ћ в€ћ
(в€’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
в€ћ в€ћ
k+1
(в€’1)k ak are
that if lim ak = 0, then the series (в€’1) ak and
kв†’+в€ћ
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.
в€ћ
n
(в€’1)n converges absolutely or condi-
29. Determine whether the series
3n
n=1
tionally.
в€ћ
(2n)!
(в€’1)n
30. Determine whether the series converges absolutely or con-
n10
n=1
ditionally.

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

в€ћ в€ћ
n n
4 4
n n+1 2
33. 34.
(в€’1) n (в€’1) n
5 5
n=1 n=1

в€ћ в€ћ
(в€’1)n (в€’1)n+1
35. 36.
n3/2 n1/2
n=1 n=1

в€ћ в€ћ
(в€’1)n (в€’1)n+1
37. ,0 < p < 1 38. ,1 < p
np np
n=1 n=1

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

3n
в€ћ в€ћ
(n + 2)2
(в€’1)n+1 (в€’1)nв€’1 2
41. 42.
(n + 1)3 n2
n=1 n=1

в€ћ в€ћ
(в€’1)n (4/3)n (в€’4)n
43. 44.
n4 (n!)n
n=1 n=1

в€ћ в€ћ
n (n + 1)!
n
(в€’1)n
45. 46.
(в€’3)
1 В· 3 В· 5 В· В· В· (2n + 1)
(2n)!
n=1 n=1

в€ћ в€ћ
2n
(n в€’ 1)
n (n!) 2
(в€’1)n+1
47. 48.
(в€’1)
n3/2
(2n)!
n=1 n=1

в€ћ в€ћ
4n 2 В· 4 В· В· В· (2n + 2)
n 2
(в€’1)n
49. 50.
(в€’1) (n!)
1 В· 4 В· 7 В· В· В· (3n + 1)
(2n)!
n=1 n=1

в€ћ в€ћ
n+1
nв€’1 5 (n + 1)
(в€’1)n+1
51. 52.
(в€’1)
24n (n + 3)
n=1 n=1

в€ћ в€ћ
n+1 (n + 2) (n + 2)
(в€’1)n
53. 54.
(в€’1)
n5/4 n7/4
n=1 n=1

в€ћ в€ћ
2
(в€’1)n
+ 2n в€’ 1)
n (3n
55. 56.
(в€’1)
2n3 n(ln n)
n=1 n=2

в€ћ в€ћ
n (ln n) (ln n)
(в€’1)n+1
57. 58.
(в€’1)
n2
n
n=2 n=1

в€ћ в€ћ p
n! nn
n
59. (в€’1) p , 0 < p < 1 60. ,0 < p < 1
(в€’1)
n n!
n=1 n=1

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

в€ћ
63. Suppose that 0 < ak for each natural number k and ak converges.
k=1
в€ћ
ap converges for every p > 1.
Prove that k
k=1

в€ћ
64. Suppose that 0 < ak for each natural number k and ak diverges.
k=1
в€ћ
ap , for 0 < p < 1.
Prove that k
k=1

65. Suppose that 0 < r < 1 and |ak+1 /ak | < r for all k в‰Ґ N . Prove that
в€ћ
ak converges absolutely.
k=1

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

8.5 Alternating Series
Deп¬Ѓnition 8.5.1 Suppose that for each natural number n, bn is positive or
negative. Then the series в€ћ bk is said to converge
k=1

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

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

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

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

Deп¬Ѓnition 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
(в€’1)k+1 ak
(a) a1 в€’ a2 + a3 в€’ В· В· В· + (в€’1) an + В· В· В· =
k=1

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

Theorem 8.5.2 Suppose that 0 < an+1 < an for all natural numbers m, and
lim an = 0. Then
nв†’в€ћ

в€ћ в€ћ
n
(в€’1)n+1 an both converge.
(a) (в€’1) an and
n=1 n=1

в€ћ n
k+1
(в€’1)k+1 an < an+1 , for all n;
an в€’
(b) (в€’1)
k=1 k=1

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

Proof.
8.5. ALTERNATING SERIES 343

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

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

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

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

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

s2n = a1 в€’ a2 + a3 в€’ a4 + В· В· В· + (в€’1)2n+1 a2n
= a1 в€’ (a2 в€’ a3 ) в€’ (a4 в€’ a5 ) в€’ В· В· В· в€’ (a2nв€’2 в€’ a2nв€’1 ) в€’ a2n < a1 .

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

lim s2n+1 = lim s2n + lim a2n+1
nв†’в€ћ nв†’в€ћ nв†’в€ћ
= lim s2n
nв†’в€ћ
= s.

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

s2n < s < s2n+1 < s2nв€’1 . . . (1)

for each natural number n. It follows that

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

and
в€ћ 2n
(в€’1)k+1 ak в€’ (в€’1)k+1 ak < a2n+1 .
k=1 k=1

Similarly,

s2n в€’ s2nв€’1 < s в€’ s2nв€’1
s2nв€’1 в€’ s2n > s2nв€’1 в€’ s
s в€’ s2nв€’1 < s2nв€’1 в€’ s2n = a2n
в€ћ 2nв€’1
k+1
(в€’1)k+1 ak < a2n .
ak в€’
(в€’1)
k=1 k=1

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

в€ћ n
(в€’1)k ak в€’ (в€’1)k ak
Part (c)
k=1 k=1

в€ћ n
(в€’1)k+1 ak в€’ (в€’1)k+1 ak
= (в€’1)
k=1 k=1
в€ћ n
k+1
(в€’1)k+1 ak < a2n+1 .
ak в€’
= (в€’1)
k=1 k=1

This concludes the proof of this theorem.
в€ћ
ak . Let
Theorem 8.5.3 Consider a series k=1

an+1
= L , lim |an |1/n = M
lim
an
nв†’в€ћ nв†’в€ћ

в€ћ
ak converges absolutely.
(a) If L < 1, then the series k=1
в€ћ
ak does not converge absolutely.
(b) If L > 1, then the series k=1
в€ћ
ak converges absolutely.
(c) If M < 1, then the series k=1
8.5. ALTERNATING SERIES 345

в€ћ
ak does not converge absolutely.
(d) If M > 1, then the series k=1
в€ћ
ak may or may not converge
(e) If L = 1 or M = 1, then the series k=1
absolutely.
в€ћ
Proof. Suppose that for a series ak ,
k=1

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