104 MATH HW 10
CLAY SHONKWILER
8.4 2. Does the following series converge or diverge?
en .
n=1
Answer: We can re-write the series as
n=1
1 e
1 e
n1
,
which is a geometric series with a = r = 1 , so the series converges to e
1 e
1
1 e
=
1 e e1 e
=
1 . e1
3. Does the following series converge or diverge?
n=1
n . n+1
Answer: Note that, by LHpitals Rule, o n 1 lim = lim = 1. n n + 1 n 1 Hence, by the nth term test, the series diverges. 6. Does the following series converge or diverge? 2 . n n n=1 Answer: We can re-write this series as
2
n=1
1 n3/2
= 2
n=1
1 n3/2
.
Since 3/2 > 1, this is a convergent p-series. 20. Does the following series converge or diverge?
n=1
1 . (ln 3)n
1
2
CLAY SHONKWILER
Answer: We can re-write this series as
n=1
1 ln 3
1 ln 3
n1
.
This is a geometric series with a = r = ln 3 1.09 > 1, so it diverges. 27. Does the following series converge or diverge?
n=1
8 tan1 n . 1 + n2
Answer: We apply the integral test:
1
8 tan1 x dx = lim b 1 + x2
1 , 1+x2
b
b
8
1 1
tan1 x dx 1 + x2
Now, letting u = tan1 x, du =
tan1 b b
so the above integral becomes
tan1 b /4 2
lim 8
/4
u2 udu = lim 8 b 2 = lim 8
b
tan1 b 2
(/4)2 2
(/2)2 (/4)2 2 2 2 2 =8 8 32 2 3 =8 32 2 3 = . 4 =8 Therefore, since 1 8 tan 2 x dx converges, the series 1+x verges, by the integral test. 39. (a): Show that
2
1
8 tan1 n n=1 1+n2
also con-
dx x(ln x)p
(p a positive constant)
converges if and only if p > 1.
MATH 104 HW 10 1 Answer: Suppose p = 1. Let u = ln x. Then du = x dx and so 2
3
dx = x ln x
ln 2
du u
b
= lim
b ln 2 b b
du u
= lim ln |u|]b 2 ln = lim [ln b ln(ln 2)] = . On the other hand, if p < 1, then again we let u = ln x, and so 1 du = x dx and
2
dx = x(ln x)p
ln 2
du up
b
= lim = lim = lim =
du
b ln 2
b ln 2 up up+1 b
p + 1 b1p 1p
b
(ln 2)1p 1p
since 1 p > 0. Finally, if p > 1, then the above computation of the integral still applies; however,
b
lim
b1p (ln 2)1p (ln 2)1p = 1p 1p 1p
dx since 1 p < 0. Therefore, we conclude that 2 x(ln x)p converges if and only if p > 1. (b): What implications does the fact in (a) have for the convergence of the series 1 ? n(ln n)p n=2
Answer: The result proved in (a) demonstrates, by the integral test, that this series converges if and only if p > 1. 8.5 2. Does the following series converge or diverge? 3 . n+ n n=1
4
CLAY SHONKWILER
Answer: Since for all n 1, Therefore,
n n, we know that n+ n n+n = 2n.
3 3 . 2n n+ n Now,
n=1
3 = 2n
n=1
3 1 2 n
3 n+ n
is a p-series with p = 1, and so diverges. Hence, since diverges as well. n 1, the series 6. Does the following series converge or diverge? n+1 . n2 n n=1
3 n=1 n+ n
3 2n
for all
Answer: Since constant terms will tend not to mean very much as n gets 1 1 big, we compare the terms of this series with n2n n = nn = n3/2 . Now,
n=1
1 n3/2
is a p-series with p = 3/2 > 1, so it converges. Hence, we expect the original n+1 1 series should converge. However, since n2 n n3/2 , a direct comparison wont demonstrate this, so we do a limit comparison instead:
n+1 n2 n lim 1 n n3/2
= lim
n3/2 (n + 1) n+1 = 1 > 0. = lim 2 n n n n n
1 n=1 n3/2 ,
n+1 Therefore, the series n2 n acts like the series n=1 say it also converges. 14. Does the following series converge or diverge?
which is to
(ln n)2 . n3/2 n=1 Answer: Since n ln grows very, very slowly, we limit compare the terms 1 of this series to the terms of the series n3/2 : n=1
(ln n)2 3/2 lim n 1 n n3/2
= lim (ln n)2 = .
n
1 Since converges, this doesnt tell us anything. However, it still seems n3/2 2 shouldnt prevent the series from converging, so lets compare the like (ln n)
MATH 104 HW 10
5 1n 5/4 :
terms of this series to the terms of another convergent series,
(ln n)2 3/2 lim n 1 n n5/4
= lim = lim
n5/4 (ln n)2 n n3/2
(ln n)2 n n1/4 By LHpitals Rule, this limit is equal to o
1 2 ln n n ln n = lim 8 1/4 . 1 1 n n n 4 n3/4
lim
Again using LHpitals Rule, this limit is equal to o
n
lim 8 1
4
1 n
1 n3/4
= lim 32
n
1 n1/4
= 0.
1 Since converges (it is a p-series with p = 5/4 > 1), this tells us that n5/4 the series (ln n)2 n3/2 n=1
converges as well. 24. Does the following series converge or diverge?
n=1
3n1 + 1 . 3n
n=1
Answer: Re-write the series as
n=1
3n1 1 + n = n 3 3
1 1 + n . 3 3
Now, 1 1 1 1 + = + 0 = = 0, 3 3n 3 3 so we see that the terms of the series do not converge to 0. Therefore, by the nth term test, the series does not converge. 38. If an is a convergent series of nonnegative numbers, can anyn=1 thing be said about (an /n)? n=1 Answer: Yes. Note that, for all n 1, an an . n Therefore, since an converges, the direct comparison test tells us that n=1
n
lim
n=1
an n
converges as well.
6
CLAY SHONKWILER
8.6 2. Does the following series converge or diverge?
n2 en .
n=1
Answer: Using the ratio test, (n + 1)2 (n + 1)2 e(n+1) = lim . n n n2 en n2 e By two applications of LHpitals Rule, this limit is equal to o 2(n + 1) 2 1 lim = lim = < 1. n n 2e 2ne e Therefore, by the ratio test, the series converges. 4. Does the following series converge or diverge? lim
n=1
n! . 10n
Answer: Again, we use the ratio test:
(n+1)! 10n+1 n! n=1 10n
lim
= lim
(n + 1)! 10n n+1 = lim = > 1, n 10n+1 n 10 n!
so we see, by the ratio test, that the series diverges. 8. Does the following series converge or diverge?
n=1
(2...