950
[ I
CHAPTER12 INFINITE
I
SEQUENCESAND SERIES
00
00
=L I
}:=a..=}:=1
12.5 AlternatingI Series
10. Lan
n=l
n=l
1. (a) An alternating series is a series whose tenDs are alternately positive and negative.
(b) An alternating
series
E~=l(
_1)n-lbn
converges
if 0 < bn+l
~ bn for all nand
n=l
n=l
(in fact the lim
lim
bn =
n-+oo
O. (Ibis
is the
11. bn
= ~ n2
n +4
Alternating Series Test.)
(c) The error involved in using the partial sum Sn as an approximation to the total sum S is the remainder
Rn = S - Sn and the size of the error is smaller than bn+l; that is,IRnI ~ bn+l. (This is the Alternating Series
(~)'=
Estimation Theorem.)
lim
n-oo
1 +-4
2 -~ 5 +~6 -~ 7 +"'=
2. -- 3
f(-1)n~
n=l
2 .Herean=(-1)n~
n +
2 .Since
n +
lim
an #
n-+oo
7
= ~
L., (-1 ),,-1 -. 4
- -4 + -4 - -4 + -4
8
lirn
9
10
11
n=1
n+ 6
b" = 0, so the series converges by the Alternating
Now bn = - 4
n+ 6
el/n
12. bn = n
. decreasmg,
.
> 0, { bn} IS
Series Test.
lim ell"
"-rxo
n=2
nn
Alternating
Series Test.
1
5. bn = ~
yn
_
is positive and {bn} is decreasing; Jim In1
nn
n-+oo
= 0, so the series converges by the
rxo
00 (-1)"-1
I=:
,,=1
yn
converges by the Alternating
14."~l (-1)"-1 (
thenf(x) =
Series Test.
6. bn= _3 1 1 >
n -
0, {bn} is decreasing, and lim b,. = 0, so the series
n-+oo
f
n=1
i=!L
3n
-
IX}
15.
7.2::: a" = 2:::(-1)" ~
2n 00
,,=1
00
1 = 2:::(-I)nbn.
n +
n=1
Now lirn b" = Jim
"-+00
n=1
~
/n
n-+oo 2 + 1 n
=
~ to.
bn
~
> 0, {b,,}
n-+OX)
n-+OX)
4 2/1n/
+
2
n
0, SO the
series
E
n=l
--
the Alternatinj
16.Sin(~) = (
is decreasing[since
=
~ n3/4
n-+oo
decreasing, an
- bn+l= ~
- 2n + 2 8n2+ 8n - 2
4n2+ 1 4n2+8n+ 5 - (4n2+ 1)(4n2+ 8n+ 5) > Oforn ~ 1],and
lim bn = lim
E
n=l
Since Jim an to
2
(in fact the limit does not exist), the series diverges by the Test for Divergence.
8. bn =
Iim b" = III
Iii
"-rxo
nn-
converges by the Alternating
1
Series Test.
00
=]
13. ~ (-1)"~
,,=2
In 1:
n
> 0, {b,,} is decreasing,and Jim b" = 0, so the series E
n-+oo
>
(~)'==
and
"-+00
I 1 . b" = _
11
4. f (-1)" _
IiI
n~
0 (in fact the limit
does not exist), the series diverges by the Test for Divergence.
3. -4
bn =
(X)
17. E(-l)"sin
"=1
(-1
)n 4
2n
2
n
converges
by the Alternating
Series
Test.
+1
Alternatively,to showthat {bn} is decreasing,we could verify that ~
dx
(4X2+1 )
'!:.=:.--
< o for x>
-
converges by t
~
E (-l)ncos
n=l
18.
1.
the Test for Di
1
OX) (_1)n+l
.
9. bn = 4 2 1 > 0, {bn} is decreasing,and Jim bn = 0, so the series E
convergesby theAlternating
n +
n-+OX)
n=l 4n2 + 1
Series Test.
19. ~n
n.
- n.n.
,-- 1 .2.
the Test for OJ
SECTION12.5 ALTERNATING
SERIES 0
00
..jii
00
= ~(-l)n~
10. ~an
951
00
lim
= ~(-l)nbn. Now limbn=
1
2 + l/Vii
n-+oo
-
1
2 f
.
O. Since
lim
n-+~
n-~
an
~
0
(in fact the limit does not exist), the series diverges by the Test for Divergence.
:he
~
(~ + 4)' -2
11.bn =
> 0 for n ? 1. {bn} is decreasingfor n ? 2 since
+ 4)(2x) - x2(3x2)
(X3
X3
-
+ 4)2
(X3
X(2X3+ 8 - 3x3) - x(8 - X3)
(X3 + 4)2
- (x3 + 4)2 < 0 "lor x > 2. Al so,
Series
/
lim bn = lim 1 14n/
n-+(X)
,limit
+
n-+(X)
(X)
=
3
O.
Thus,
the
series
~
n
( -1
-;-
)n+1
L..,
n
n=1
2
4
converges
by
the
Alternating
Series
Test.
+
-
12. bn = e1/n > 0 for n ? 1. {bn} is decreasing
since
(- )n' = x . e1/Z(-1/x2)2 - e1/z. 1 = -e1/Z(13 + x) < 0 for x > O.Also, lim bn= 0 since
e1/Z
x
x
X
(X)
"-+(X)
l
n=2
E
n
n
(X)
14.
"-+(X)
n
(- )
lnn
-1
(-1)"
,,=1
n-+CX)
cosn7r
~ cosn7r
Z-+(X)
~=
n
L.,
n=l
~
lnn
-.
lux
H
lim
z-+(X)
~
n
bn = -Inn > Oforn?
.
2, and Jff(x)
n
=-,
lnx
x
e, so {bn} is eventuallydecreasing.Also,
lim-=
n =
= 00, sothe seriesdivergesby the Testfor Divergence.
X
(n )
,,=2
b
1/
_1
Z-+(X)
(-l)n-
z-,oo x
~.
n
convergesby the AlternatingSeriesTest.
lim
x
1
E
< 0 for x>
(-l)n
~00 (-l)n
00
. L.,
~
=
15.E~
n=l
nn3/4
n=l
n~l~=n~l~.
n
(X)
I-hlx
lim bn = lim
= lim _I x
n
= 0+
n
n-+(X)
15
_I n
. lim
n
thenf(x) =
tating
~n
lim ell" = 1. Thus,the series~(-1)"-1
L..,
n=1
13. E(-l)"-
n-+(X)
1/n
1.
IS
.!1=
1
d
= 0, so the series converges by the Alternating
.
. .
d
d
I
+
.
ecreasmg
an
1m
~
andposItIve
positive an
and n-+(X)
lim
decreasing
n-+CXI nn
1
Series Test.
0
/ 4 ==
.
0,,soSOthe
the senes
series converges
converges
by
by
the
theAlternating
AlternatingSeries
SeriesTest.
Test.
= 0 ifn
16. sin(~ 27r)
is even and (-l)k
decreasing, and lim
(2 1 1)1
n+
n-+(X)
17. E
17.
(-l)n
sin~.
bn
n=1
=
sin
n
= 2k +
ifn
E
1, so the seriesis n=O
( 2n+1.
(-l)n
)
,.
bn =
( 2n+1. 1
= 0,
so the series converges by the Alternating
Series Test.
0 for
n
sin
)
, >
0,
{bn}
is
.
~
>
?:
2 and
sin
n
~
?: sin
n
--.:!!:.--,
and
lim
n + 1
n-+(X)
~
=
sin
0
=
0, so the
series
n
converges by the Alternating Series Test.
18.
18.
E(-l)n
cos
n=l
(~ ). lim
n
cos
n-+(X)
(~
)
=
cos(O)
=
1, so
n
lim
n-+(X)
(-l)n
cos
( ~ )
n
does
not
exist
and
the
series
diverges
by
the Test for Divergence.
nating
n
19.-n , = n.n.
1 2
n.
..
n
-n n, =
n
?: n
. . . . n
the Test for Divergence.
:::}
Ii n
lim
m-=oo~
n-(X)
n-~
n.
n!
(- l )n nn
"
00
:::}
I
"
lim
1m
n-+(X)
n-~
n.
,
does not exist. So
Sothe
theseries
seriesdiverges
divergesby
by
.,
952 0
CHAPTER12 INFINITESEQUENCES
AND SERIES
Ii
( n)n
divergesby the Test for Divergence since n-+~ "5 =00
20. f
(-i)n
n=l
21.
1
( )
_~
=>
lim
n-+oo
doesnot exist.
n
5
~
(X)
25. The series
n=]
{s.}.. . . . . . .
becausebn+l
...
0
2n
lim-=n-+(X)n!
10
{a.}.
2
n
b8 = 28/8! ~
is less than the
-1
By the Alternating Series Estimation Theorem, the error in the
L
(X)
26. The series
n=l
Is - slol :$ b11= 1/{11)3/2 ~ 0.0275(to four decimalplaces.
bn+l
n+]
~
=
rounded up).
b5 = 5/45 ~ I
22.
Theorem,
n=
1
. . . . {So}
. . . . .
0
.
-0.2
27.
{an}
10
.
By the Alternating Series Estimation Theorem, the error in the
00 (-1)"-1
approximation E
,,=1
n
3
Is- slol $ bu = 1/113
b7
= 751 =
'1
00
(_l)n+l
E
n=l
n
not
changethe
5
6
za.b6=86=2
~~~
~ 0.90112 is
L..,
n=l
8n
change
the
~ 0.0007513.
fou
72
23. The series f(-l)n-l~
satisfies
(i) of theAlternatingSeries
Testbecause
~
n=l
(ii) nl~~
~
= 0, so the series is convergent. Now bIG =
by the Alternating Series Estimation Theorem, n
~=
0.01 and bll
=
~=k
~ ~
~ = 0,
(ii) lim
n-+oo
b6
n
= 1/64
(_l)n+1
~
0.008
<
0.01, so
= 10. (That is, since the 11th term is less than the desired error,
satisfies (i) of the Alternating Series Test because <n-l1F
<
so the seriesis convergent.Now bs = 1/54 = 0.0016 > 0.001 and
~ 0.00077
107
00
we need to add the first 10 terms to get the sum to the desired accuracy.)
24. The series
29.b7=-=t
< ~ and
< 0.001, so by the AlternatingSeriesEstimationTheorem,n = 5.
~
and
L
n=l
(_l)n-lr
Ion
Adding
b7 to
places.
is 0.06~
1
36.6!
JO.b6=-=
~i=!2:~
L.., 3nn!
n=l
change
the
foul
s