Page 277 - Calculus Workbook For Dummies

261
Chapter 13: Infinite Series: Welcome to the Outer Limits

s ! n n converges.
3
n 1 ! n
=
Try the ratio test:
n 1
+
^ n + 1h
n 1 n 1 n 1 1
+
+
+ -
^ n + 1h ! ! n n + 1h ^ n + 1h ^ n + 1h
^
lim = lim = lim = lim = 0
n ^ n + 1h !n ^ n + 1h n n
n " 3 n n " 3 n n " 3 n n " 3 n
! n
By the ratio test, the series converges.
3
t ! ! n n diverges.
=
n 1 4
^ n + 1h !
+
4 n 1 ^ n + 1h !4 n n + 1
Try the ratio test: lim = lim n 1 = lim = 3
+
n " 3 ! n n " 3 ! n 4 n " 3 4
4 n
Thus the series diverges.
3 n 1 n + 1 n + 1 1
+
u !^ - 1h diverges. This one is a no-brainer, because lim = , the first condition of
n 1 3 n + 1 n " 3 3 n + 1 3
=
the alternating series test is not satisfied, which means that both the alternating series and the
series of positive terms are divergent.
3
n n + 1
*v !^ - 1h 2 diverges.
=
n 3 n - 2
Check the two conditions of the alternating series test:
n + 1
1. lim
2
n " 3 n - 2
1
= lim (L’Hôpital’s Rule)
n " 3 2 n
= 0 Check.
2. Are the terms non-increasing?
1 +
n + 1 $ ^ n + h 1 ?
2
n - 2 ^ n + h 2 2
1 -
n + 1 $ n + 2 ?
2
2
n - 2 n + 2 n - 1
2
2
2 _
^ n + h n + 2 n - i n + h n - 2i ?
1 $ ^
1 _
2
2
2
3
3
n + 2 n - n + n + 2 n - 1 $ n - 2 n + 2 n - 4 ?
3
3
2
2
n + 3 n + n - 1 $ n + 2 n - 2 n - 4 ?
2
n + 3 n + 5 $ 0 ? Check .
Thus the series is at least conditionally convergent. And it is easy to show that it is only
conditionally convergent and not absolutely convergent by the direct comparison test. Each
3
term of ! n + 1 has a larger numerator and a smaller denominator — and is thus greater than
2
2
n -
3
the corresponding term of ! n 2 . ! n 2 is the same as the divergent harmonic series, ! , and
3
3
1
=
n 3
n
n
n 3
=
3
=
n 3
=
therefore ! n + 1 is also divergent. n n 3
2
=
n 3 n - 2