Page 274 - Calculus Workbook For Dummies

258 Part IV: Integration and Infinite Series

3
*l ! 1 diverges.
n 2 n ln n + sinn
=
1. You know you can integrate # 1 with a simple u-substitution, so do it, and then you’ll be
x lnx
able to use the integral comparison test.
3
# dx
x lnx u = lnx when x = 2 , u = ln2
2
c du = dx when x = , c u = lnc
= lim # dx x
c " 3 x lnx
2
ln c
= lim # du
u
c " 3
ln 2
6
= lim lnu@ ln c
c " 3 ln 2
= lim ln lnc - ln ln2hj
^
^
h
`
c " 3
= 3
By the integral comparison test, ! 1 diverges with its companion improper integral,
3
=
n 2 n lnn
3
# dx .
x lnx
2
1
2. Try the direct comparison test. Won’t work yet because is sometimes less than
n lnn + sinn
1 .
n lnn
3. Try multiplication by a constant (always easy to do and always a good thing to try).
3
3
! 1 diverges, thus so does ! 1 = ! 1 .
3
1
n 2 n lnn 2 n 2 n lnn n 2 2 n lnn
=
=
=
1
4. Now try the direct comparison test again. It’s easy to show that is always greater
n lnn + sinn
3
1 2h, and thus the direct comparison test tells you that ! 1
than ^ n $ must
2 n lnn n 2 n lnn + sinn
=
3
diverge with ! 1 .
n 2 2 n lnn
=
m ! n 2 3 converges.
3
n 1 e n
=
This is ready-made for the integral test:
3
3 2 c 2 c 3
# x dx = lim# x dx = lim # du u = 1 lim - e A c = - 1 lime 1 3 - 1 o = 1
1
7
u
-
e x 3 c " 3 e x 3 c " 3 3 e 3 c " 3 1 3 c " 3 e c e 3 e
1 1 1
Because the integral converges, so does the series.
3
*n ! n 3 converges.
n 1 ! n
=
3
1. Try the limit comparison test with the convergent series, ! 1 , as the benchmark.
n 1 ! n
=
n 3
! n ! n n 3
lim = lim = 3 No good. This result tells you nothing.
n " 3 1 n " 3 ! n
! n