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= 11
n· 2 n
converges.- Consider the integral
∫∞0x
ex
dx=
∫∞0xe−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,
∫∞0xe−xdx=klim→∞∫k0xe−xdx=klim→∞
[
−xe−x−e−x]k
0 =klim→∞
[
−ke−k−e−k+ 1]k
0 =1. Since the improper
integral converges,
∑∞
n= 0n
enconverges.- Use the limit comparison test, comparing
to the series
∑∞
n= 11
n
, which is known todiverge. 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= 11
n
diverges, so theseries
∑ 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= 1n
(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−^1diverges,∑∞
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= 1n+ 1
7 n^2 − 5. Applying the informal
principle, you have the series 7∑ 1
n.
Apply the limit comparison test and obtainnlim→∞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
nis a
harmonic series and it diverges, the series
∑∞
n= 1n+ 1
7 n^2 − 5
diverges. Therefore, the series
∑∞
n= 1(− 1 )n
n+ 1
7 n^2 − 5does 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 havef′(x)=
−(7x^2 + 14 x+5)
(7x^2 −5)^2
,x/=±√
5
7.
Note thatf′(x)<0 forx≥1 and