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Кузбасский государственный технический университет

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[НЕСОРТИРОВАННОЕ]

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8.5. ALTERNATING SERIES		341
	∞
	Xk
63. Suppose that 0 < ak for each natural number k and		ak converges.
	=1
	∞
	Xk
Prove that	akp converges for every p > 1.
	=1
	∞
	Xk
64. Suppose that 0 < ak for each natural number k and		ak diverges.
	=1
	∞
	Xk
Prove that	akp, for 0 < p < 1.
	=1

65.	Suppose that 0 < r < 1 and \|ak+1/ak\| < r for all k ≥ N. Prove that
	∞
	Xk
	ak converges absolutely.
	=1
	∞	n	an
	Xk	(−1)
66.	Prove that	(−1)	3 + bn converges absolutely if 0 < a < b.
	=1

8.5Alternating Series

Deﬁnition 8.5.1 Suppose that for each natural number n, bn is positive or

P∞

negative. Then the series k=1 bk is said to converge

(a)	absolutely if the series Pk∞=1 \|bk\| converges;
(b)	conditionally if the series Pk∞=1 bk converges but Pk∞=1 \|bk\| diverges.
Theorem 8.5.1 If a series converges absolutely, then it converges.
	∞	\|bk\| converges. For each natural number k, let
Proof. Suppose that		\|bk\| converges. For each natural number k, let
	=1
	Xk
ak = bk + \|bk\| and ck = 2\|bk\|. Then 0 ≤ ak ≤ ck for each k. Since
	∞	∞	∞
	X	X	Xk	\|bk\|,
	k=1	ck =	2\|bk\| = 2	\|bk\|,
	k=1	k=1	=1

342		CHAPTER 8.	INFINITE SERIES
∞		∞
X		Xk
the series	ck converges. by the comparison test		ak also converges. It
k=1		=0
follows that
	∞	∞
	Xbk = X(ak − \|bk\|)

k=1 k=1

∞∞

		X	X
	=	ak −	\|bk\|
		k=1	k=1
	∞
and the series	X
and the series	bk converges. This completes the proof of the theorem.
	k=1

Deﬁnition 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
(a) a1 − a2 + a3 − · · · + (−1)n+1an + · · · =	Xk
(a) a1 − a2 + a3 − · · · + (−1)n+1an + · · · =	(−1)k+1ak
	=1
	∞
(b) −a1 + a2 − a3 + · · · + (−1)nan + · · · =	Xk
(b) −a1 + a2 − a3 + · · · + (−1)nan + · · · =	(−1)kak.
	=1

Theorem 8.5.2 Suppose that 0 < an+1 < an for all natural numbers m, and

lim an = 0. Then

n→∞

∞∞

(a)(−1)nan and (−1)n+1an both converge.

n=1 n=1

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

		∞	∞
(c)			(−1)kak − (−1)kak		< an+1, or all n.
	k=1		k=1
		X	X

Proof.

8.5. ALTERNATING SERIES	343
	n
	Xk
Part (a) For each natural number n, let sn =	(−1)k+1ak. Then,
	=1
2n+2	2n

s2n+2 − s2n =	(−1)k+1ak − (−1)k+1ak
k=1	k=1

=(−1)2n+3a2n+2 + (−1)2n+2a2n+1

=a2n+1 − a2n+2 > 0.

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

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

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

s2n = a1 − a2 + a3 − a4 + · · · + (−1)2n+1a2n

= a1 − (a2 − a3) − (a4 − a5) − · · · − (a2n−2 − a2n−1) − a2n < a1.

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

lim s2n+1	= lim s2n	+ lim a2n+1
n→∞	n→∞	n→∞
	= lim s2n
	n→∞
	= s.
It follows that
	lim sn = s
∞	n→∞		∞
∞			∞
and the series (−1)n+1ak converges to s and the series			(−1)nak con-
n=1			n=1
X			X
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

∞ (−1)k+1ak − 2n (−1)k+1ak

< a2n+1.

k=1

Similarly,

s2n − s2n−1 < s − s2n−1

s2n−1 − s2n > s2n−1 − s

s − s2n−1 < s2n−1 − s2n = a2n

∞ (−1)k+1ak − 2n−1(−1)k+1ak

< a2n.

k=1

It follows that for all

natural numbers n,

∞ (−1)k+1ak − n

(−1)k+1ak

< an+1.

k=1

∞

Part (c) k=1(−1)kak − k=1(−1)kak

∞

(−1) (

k=1

(−1)k+1ak − (−1)k+1ak)

k=1

∞

k+1a

< a

(−1)

k −

(−1)

2n+1

k=1

This concludes the proof of this theorem.

P∞

Theorem 8.5.3 Consider a series k=1 ak. Let


lim	an+1	=	L ,	lim	a	n\|	1/n = M
lim	an		L ,	lim	a		1/n = M
n→∞	an			n→∞	\|

(a)	If L < 1, then the series	∞	ak	converges absolutely.
		k=1
		Pk∞
(b)	If L > 1, then the series P∞=1 ak			does not converge absolutely.
(c)	If M < 1, then the series Pk=1 ak			converges absolutely.

P∞

k=1 ak may or may not converge

lim |an|1/n = M.

n→∞

8.5. ALTERNATING SERIES

345

P∞

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

(e) If L = 1 or M = 1, then the series absolutely.

	∞
Proof. Suppose that for a series	Xk
Proof. Suppose that for a series	ak,
	=1

an+1

lim = L and

n→∞ an

				∞
				Xk
Part (a) If L < 1, then the series				\|ak\| converges to the ratio test, since
n→∞ \|				=1
n→∞ \|	an	\|		n→∞	an
lim	an+1		= lim		an+1	= L < 1.
	\| \|

∞

Hence, the series ak converges absolutely.

k=1

∞

Part (b) As in Part (a), the series |ak| diverges by the ratio test if L > 1,

k=1

since

n→∞ |

n→∞

lim

an+1

= lim

an+1

= L > 1.

∞

nlim |ak|1/n

= M < 1.

|ak| converges by the root test, since

Part (c) If M < 1, then the series

→∞

∞

|ak| diverges by the root test as in

Part (d) If M > 1, then the series

Part (c).

∞

Part (e) For the series

and

, L = M = 1, but

diverges

∞

k=1

and

converges by the p-series test. Thus, L = 1 and M = 1 fail to

determine convergence or divergence.

346	CHAPTER 8. INFINITE SERIES

This completes the proof of Theorem 8.5.3.

Exercises 8.3 Determine the region of convergence of the following series.

71.

∞

(−1)nxn

72.

∞

(−1)n(x + 2)n

n=1

3nn2

73.

∞

(−1)n(x − 1)n

74.

∞

(−1)nn!(x − 1)n

n=1

∞

(x + 2)n

75.

(−2)nxn

76.

n=0

n=1

2nn2

77.

∞

(

1)n (x + 1)n

78.

∞

(−1)n(x − 3)n

−

n=1

3nn3

n=1

n3/2

79.

∞

(2x)n

80.

∞

(−1)nxn

n=1

(2n)!

81.

∞

(n + 1)!(x − 1)n

82.

∞

(−1)n(2n)!xn

n=1

83.

∞

n2(x + 1)n

84.

∞

(−1)nn!(x − 1)n

· · ·

(2n + 1)

n=1

85.

∞

(−1)n(n!)2(x − 1)n

86.

∞

(−1)n3nxn

n=1

3n(2n)!

n=1

23n

87.

∞

(−1)n(x + 1)n

88.

∞

ln(n + 1)2n(x + 1)n

n=1

(n + 1) ln(n + 1)

n=1

n + 2

89.

∞

(−1)n(ln n)3nxn

90.

∞

(−1)n1 · 3 · 5 · · · (2n + 1)

4nn2

· · ·

(2n + 2)

n=1

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