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8.4. SERIES WITH POSITIVE TERMS

327

Prove that ( n

rk)∞

converges if and only if |r| < 1.

n=1

∞

Prove that (

)

diverges.

n=1

Prove that (

)

∞

diverges.

k ln k

n=2

10. Prove that for each natural number m ≥ 2,

(a)

Z1m(ln t)dt < ln(m!) < Z1m+1

(ln t)dt

(b)

m(ln(m) − 1) < ln(m!) < (m + 1)(ln(m + 1) − 1).

(c)

(m + 1)m+1

< m! <

em−1

(d)

lim

(m!)1/m = + .

m→+∞

∞

(e)

lim

(m!)1/m

m→+∞

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

sin(1/n)

∞

12. Prove that

n=1 converges to 1.

(1/n)

sin n ∞

13. Prove that

n=1 converges to zero.

8.4 Series with Positive Terms

Theorem 8.4.1 (Algebraic Properties) Suppose that			∞	a		and	∞	b
are convergent series and c > 0. Then			Pk=1		k		Pk=1		k
∞	∞	∞

XX X

(i)	(ak + bk) =	ak + bk
k=1	k=1	k=1

328		CHAPTER 8. INFINITE SERIES
∞	∞	∞

XX X

(ii)	(ak − bk) =	ak − 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

∞∞

X	X
ck and	ck
k=1	k=m
either both converge or both diverge.
Proof.
Part (i)

∞

(ak ± bk) = nlim		(ak ± bk)
k=1	→∞	k=1	±	n→∞		k!
=	n→∞	k!	±	n→∞	n	k!
		n			n
	lim	Xk		lim	X
	lim	a		lim	b
	∞	=1			k=1
	∞	∞
	X	X
=	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
k=1	n→∞ k=1
		n→∞	k!
			n
	= c	lim	Xk
	= c	lim	a
		∞	=1
		∞
	= c	Xk
	= c	ak.
		=1

8.4. SERIES WITH POSITIVE TERMS

329

Part (iv) We observe that

∞ m−1 ∞

X X X

ak = ak + ak.

k=1 k=1 k=1

Therefore,

∞n

X		X
	ak = lim	ak
k=1	n→∞ k=1
	m−1	n
	X	X
	=	ak + lim	ak.
	k=1	n→∞ 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 theorem.

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							kn=1 ak ≤ M, for all natural numbers
	n	, then	∞	a		If	there exists no such M, then the series
		, then	k=1	k converges.		If	P
	diverges.P		converges, then Pk∞=1 ak
(b)	If Pk∞=1 bk		converges, then Pk∞=1 ak					converges.
(c)	If Pk∞=1 ak		diverges, then Pk∞=1 bk diverges.
(d)	If cn > 0 for all natural numbers n, and
				nlim		cn	= L, 0 < L < ∞,
				nlim		a	= L, 0 < L < ∞,
				→∞		n
	then the series Pk∞=1 ak and				Pk∞=1 ck either both converge or both diverge.

330	CHAPTER 8. INFINITE SERIES

Proof. Let An =	ak, Bn = bk, 0 < an ≤ bn for all natural numbers
k=1	k=1

n. The sequences {An}∞n=1 and {Bn}∞n=1 are strictly increasing sequence. Let A represent the least upper bound of {An}∞n=1 and let B represent the least upper bound of {Bn}∞n=1

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

		∞
	A =	Xk
	A =		ak.
		=1
∞		∞
X	Xk
Part (b) If	bk converges, then	bk = B and An ≤ Bn ≤ B for all
k=1	∞	=1
	∞
	Xk
natural numbers n. By Part (a),		ak	converges to A.

∞

Part (c) If ak diverges, then the sequence {An}∞n=1 diverges. Since {An}∞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}∞n=1 diverges.

Part (d) Suppose that 0 < an and 0 < cn, 0 < L < ∞, = L2 and

lim cn = L.

n→∞ an

Then there exists some natural number m such that

	cn	− L	<	1
an		− L	<	2

8.4. SERIES WITH POSITIVE TERMS

331

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

−

− L <

an ≤ cn ≤

L an.

k=m ak ≤ k=m mck

L k=m ak

≤ 2

∞

L m

∞

If (

ak)

diverges, then (

ak) diverges and, hence (

ck)

k=m

n=1

k=m

n=m

and ( n

ck)∞

both diverge.

k=1

∞

If (k=1 ak)k=1 converges, then ( 2 L k=m ak)n=m converges and, hence,

(

)∞

( n

)∞

	ck	and	ck	both converge.
k=m	n=m	k=1		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

lim	an+1	= r.

n→∞	an
Then the series Pk∞=1 ak

(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

lim an+1 = r.

n→∞ an

332	CHAPTER 8. INFINITE SERIES

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

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

+ r

am+k < (r + )kam =

am . . .

(2)

Hence, by (2), we get

∞

m−1

∞

an =

an +

am+k

n=1

m−1

∞

+ r

< n=1 an + k=0

m−1

an +

−

1+r

m−1

n=1

−

an +

2am

n=1

< ∞.

∞

It follows that the series

an converges.

n=1

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

an < 3r − 1an < an+1

for all n ≥ m. It follows that

0 < am ≤ lim am+k = lim an.

k→∞ n→∞

P∞

k=1 ak

8.4. SERIES WITH POSITIVE TERMS

333

∞

By the Divergence test, the series an diverges.

∞

Part (c) For both series

n=1


1		∞		1
	and	X			,
n	and	n=1		n2	,
		n=1
	lim	an+1	= 1.
	lim		= 1.
n→∞		an

∞	1		∞	1
X			X
But, by the p-series test,	n	diverges and	n=1	n2	converges. Thus, the
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→∞

Then the series

(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

r < (an)1/n <			r + . . .	(3)
−	(an)1/n	− r	<
−
for all natural numbers n ≥ m.

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

334

CHAPTER 8.

INFINITE SERIES

it follows that

∞

m−1

∞

ak =

an +

n=1

m−1

∞

+ r

< n=1 an + n=m

m−1

1 + r

n=1

−

an +

1+r

m−1

+ r

= n=1 an +

2 r

−

< ∞.

∞

Therefore, ak converges.

n=1

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

1 < 1 + r = r + < (an)1/n 2

1 < 1 + r n < an.

										∞
										X
It follows that nlim an 6= 0 and, by the Divergence test, the series											an
→∞										n=1
diverges.
		∞	1		∞		1
		X		and	X				we have r = 1, where
Part (c) For each of the series			n	and	n=1	n2			we have r = 1, where
		n=1			n=1
		r = lim (an)1/n.
		n→∞
∞	1				∞		1
X					X
But the series	n	diverges and the series						n2		converges by the p-series
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

Deﬁne what is meant by

∞

ak.

Deﬁne what is meant by the sequence of nth partial sums of the series

∞

ak.

k=1

∞

Suppose that a 6= 0. Prove that Xark

converges to

if |r| < 1.

1 − r

k=0

∞

Prove that the series

converges to

k(k + 2)

∞

−

Prove that

converges to

if p > 1 and diverges otherwise.

∞

Prove that

n + 1

n=1 is an increasing sequence and the series n=1 ln

n + 1

diverges.

∞

(−1)kxk converges to

if |x| < 1.

Prove that

1 + x

∞

x2k converges to

−

if |x| < 1.

Prove that

∞

(−1)kx2k converges to

if |x| < 1.

Prove that

1 + x2

∞

10. Prove that if ak converges, then lim ak = 0. Is the converse true?

k→∞

k=0

Explain your answer.

∞		∞
X		Xk
11. Suppose that if	ak converges to L and	bk converges to M. Prove
k=0		=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

∞

(c)(ak − bk) converges to L − M.

k=0

∞

(d)akbk may or may not converge to LM.

k=0
∞	Z1
X		∞ t1p dt converges. Deter-
12. Prove that k=1 k1p converges if and only if		∞ t1p dt converges. Deter-

mine the values of p for which the series converges.

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

		∞
		Xk
Let ak = f(k) for each natural number k. Then the series			ak	con-
		=1
verges if and only if Za∞ f(x)dx converges.
			n
14. Suppose that 0 ≤ ak ≤ ak+1 for each natural number k, and sn =			Xk
				ak.
			=1
			∞
Prove that if sn ≤ M for some M and all natural numbers n, then				ak
converges.			=1
converges.			Xk
15. Suppose that 0 ≤ ak ≤ bk for each natural number k. Prove that
∞		∞
X		Xk
(a) if	bk converges, then	ak converges.
k=1		=1
∞		∞

(b)	if	ak diverges, then	bk diverges.
	k=1		k=1
		∞
		Xk	ak diverges.
(c)	if klim ak 6= 0, then		ak diverges.
	→∞	=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	Xk
1, then	ak diverges.
	=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 < klim (ak/bk) < ∞, then	∞	ak converges if and only if	∞
if 0 < klim (ak/bk) < ∞, then		ak converges if and only if	bk
→∞	k=1		=1
converges.	X		Xk

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	Xk
1, then	ak	diverges.
	=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.
=1
Xk
∞		∞
23. A series	ak is said to converge absolutely if	\|ak\| converges. Sup-
k=1		=1
X		Xk

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
	∞			∞
	X	1/k		Xk
24. A series	ak is said to converge absolutely if			\|ak\| converges. Sup-
	k=1			=1
pose that klim (\|ak\|)			= p. Prove that
	→∞

∞

(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

converges absolutely,

then it converges.

Is the

converse true? Justify your answer.

, b

lim

= p.

∞

26. Suppose that

k 6= 0

k 6= 0 for any natural number

and k→∞

ak converges absolutely

Prove that if 0 < p < 1, then the series

∞

and only if

bk converges absolutely.

∞

27. A series

ak is said to converge conditionally if

converges but

k=1

8.4. SERIES WITH POSITIVE TERMS				X		339
Xk	\|	k\|		X	n
∞	a		diverges. Determine whether the series	∞	(−1)n+1	converges
=1	a		diverges. Determine whether the series	n=1		converges
=1				n=1
conditionally or absolutely.

28.Suppose that 0 < ak and |ak+1| < |ak| for every natural number k. Prove

∞∞

		X	k+1a	Xk	ka
that if lim	a	k=1(−1)	k+1a	=1(−1)	ka
k→+∞	k = 0, then the series	k=1(−1)	k and	=1(−1)	k are

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.

∞	(−1)	n n
29. Determine whether the series				converges absolutely or condi-
29. Determine whether the series			3n	converges absolutely or condi-
n=1
X
tionally.
∞	n (2n)!
X	(−1)
30. Determine whether the series	(−1)		n10		converges absolutely or con-
n=1
ditionally.

In problems 31–62, test the given series for convergence, conditional convergence or absolute convergence.

∞

31. X(−1)n n!

n=1

∞n

33.X(−1)nn 54

n=1

35.	∞		(−1)n
	X
	n=1		n3/2
	n=1
37.	∞	(−1)n , 0 < p < 1
	X
	n=1		np
	n=1
	∞		n (n + 1)
39.		(−1)
39.		(−1)		n2 + 2

n=1

∞

n+1 5n

32.

(−1)

n=1

34.

∞

(−1)n+1n2

n=1

36.

∞

(−1)n+1

n=1

n1/2

38.

∞

(−1)n+1 , 1 < p

n=1

∞

n+1 (n + 1)2

40.

(−1)

n=1

340

∞

(n + 2)2

41.

(−1)n+1

n=1

(n + 1)3

43.

∞

(−1)n(4/3)n

n=1

∞

45.

(−3)

n=1

(2n)!

∞

n (n!)22n

47.

(−1)

n=1

(2n)!

∞

49.

(−1)

(n!)

n=1

(2n)!

∞

5n+1

51.

(−1)n−1

n=1

24n

∞

n+1 (n + 2)

53.

(−1)

n=1

n5/4

55.

∞

( 1)n (3n2 + 2n − 1)

−

n=1

2n3

∞

n (ln n)

57.

(−1)

n=2

∞

n n!

59.

(−1)

, 0 < p < 1

n=1

∞

n n!

61.

(−1)

, 1 < p

n=1

CHAPTER 8. INFINITE SERIES

∞3 n

n−1

42.

n=1

(−1)

44.

∞

(−4)n

n=1

(n!)n

∞

(n + 1)!

46.

(−1)n

· · ·

(2n + 1)

n=1

48.

∞

(

1)n+1 (n − 1)

−

n=1

n3/2

∞

50.

(

−

1)n

2 · 4 · · · (2n + 2)

· · ·

(3n + 1)

n=1

∞

n+1 (n + 1)

52.

(−1)

n=1

(n + 3)

∞

n (n + 2)

54.

(−1)

n=1

n7/4

56.

∞

(−1)n

n=2

n(ln n)

∞

(ln n)

58.

(−1)n+1

n=1

∞

n np

60.

(−1)

, 0 < p < 1

n=1

∞

n np

62.

(−1)

, 1 < p

n=1

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