366 STEP 4. Review the Knowledge You Need to Score High
- By the ratio test, limn→∞
1
(n+1)· 2 n+^1
·
n· 2 n
1
=nlim→∞
n
2(n+1)
=
1
2
; therefore,
∑∞
n= 1
1
n· 2 n
converges.
- Consider the integral
∫∞
0
x
ex
dx=
∫∞
0
xe−xdx. Integrate by parts, with
u∫=x,du=dx,dv=e−xdx, andv=−e−x.
xe−xdx=−xe−x+
∫
e−xdx=−xe−x−e−x+C. Therefore,
∫∞
0
xe−xdx=
klim→∞
∫k
0
xe−xdx=klim→∞
[
−xe−x−e−x
]k
0 =klim→∞
[
−ke−k−e−k+ 1
]k
0 =1. Since the improper
integral converges,
∑∞
n= 0
n
en
converges.
- Use the limit comparison test, comparing
to the series
∑∞
n= 1
1
n
, which is known to
diverge. Divide
n+ 1
n(n+2)
÷
1
n
=
(n+1)/n
n/(n+2)
=
n+ 1
n+ 2
. The limit
nlim→∞
n+ 1
n+ 2
=1, and
∑∞
n= 1
1
n
diverges, so the
series
∑ n+ 1
n(n+2)
diverges.
- Use the ratio test.
nlim→∞
[
n+ 1
(n+2)n+^1
·
(n+1)n
n
]
=
nlim→∞
[
(n+1)n+^1
(n+2)n+^1
·
1
n
]
=0; therefore,
∑∞
n= 1
n
(n+1)n
converges.
6.Use the ratio test,
nlim→∞
∣∣
∣
∣
(−1)n
(n+1)!
·
n!
(−1)n−^1
∣∣
∣
∣=
nlim→∞
∣∣
∣∣ −^1
n+ 1
∣∣
∣∣=lim
n→∞
1
n+ 1
=0, so the series
∑∞
n= 1
(−1)n−^1
n!
converges absolutely.
7.
∑∞
n= 1
(−1)n−^1
n+ 1
n
=
∑∞
n= 1
(−1)n−^1
(
1 +
1
n
)
=
∑∞
n= 1
(
(−1)n−^1 +
(−1)n−^1
n
)
=
∑∞
n= 1
(−1)n−^1 +
∑∞
n= 1
(−1)n−^1
n
. Since
∑∞
n= 1
(−1)n−^1
diverges,
∑∞
n= 1
(−1)n−^1
n+ 1
n
diverges.
8.Begin by inspecting the absolute values of
∑∞
n= 1
∣∣
∣∣(− 1 )n n+^1
7 n^2 − 5
∣∣
∣∣which is equivalent to
∑∞
n= 1
n+ 1
7 n^2 − 5
. Applying the informal
principle, you have the series 7
∑ 1
n
.
Apply the limit comparison test and obtain
nlim→∞
n+ 1
7 n^2 − 5
1
n
=nlim→∞
n+ 1
7 n^2 − 5
·
n
1
=
nlim→∞
n^2 +n
7 n^2 − 5
=
1
7
. Since
∑∞
n= 1
1
n
is a
harmonic series and it diverges, the series
∑∞
n= 1
n+ 1
7 n^2 − 5
diverges. Therefore, the series
∑∞
n= 1
(− 1 )n
n+ 1
7 n^2 − 5
does not converge
absolutely. Next examine the alternating
series
∑∞
n= 1
(−1)n
n+ 1
7 n^2 − 5
. Let
f(x)=
x+ 1
7 x^2 − 5
, and you have
f′(x)=
−(7x^2 + 14 x+5)
(7x^2 −5)^2
,x/=±
√
5
7
.
Note thatf′(x)<0 forx≥1 and