Math1024 Answer to Homework 8
Exercise 4.2.20 (2)
p
p
nn
n
limn→∞ |an | = limn→∞
= 0. The series diverges.
log n
Exercise 4.2.20 (4)
(
q
p
ap , a ≥ b
p
limn→∞ n |an | = limn→∞ n (an (1 + ( ab )n ))p = limn→∞ ap (1 + ( ab )n ) n = p
b , b>a
p
Thus when a ≥ b, the series converges if a < 1 and when a < b, the series converges if bp < 1.
Exercise 4.2.20 (10)
p
a 2n−n2
1
1
= a < 1. The series converges.
limn→∞ n |an | = limn→∞ (1+ ) n = limn→∞
a
n
e
(1 + )n−2
n
Exercise 4.2.21(3)
We need to use Raabe test here to determine the convergence of the series:
lim n(1 − |
n→∞
2n
2
an+1
|) = lim
= <1
n→∞ 3n + 4
an
3
.
So the series diverges.
Exercise 4.2.22 (5)
a1 + b 1 n + c 1 n 2
c1
an+1
| = lim
=
n→∞ a2 + b2 n + c2 n2
n→∞ an
c2
c1
c1
Thus the series converges when
< 1 and diverges when
> 1.
c2
c2
Exercise 4.2.23 (3)
(n + 1)(2n + 2)(2n + 1)
4
an+1
=
r →
r and the ratio test, the series converges for
By
an
(3n + 3)(3n + 2)(3n + 1)
27
an+1 27
27
27
= (n + 1)(2n + 2)(2n + 1) 27 >
|r| <
and diverges for |r| > . Moreover, if |r| = , then 4
4
4
an (3n + 3)(3n + 2)(3n + 1) 4
1. Therefore |an | is increasing and cannot converge to 0. Therefore the series also diverges.
27
We conclude that the series converges if and only if |r| < .
4
Exercise 4.2.23 (8)
an+1
(2n + 2)(2n + 1)n2n
4
e2
By
=
r
→
r
and
the
ratio
test,
the
series
converges
for
|r|
<
an
(n + 1)2n+2
e2
4
2
e
e2
and diverges for |r| > . Moreover, if |r| = , then
4
4
e2
We conclude that the series converges if and only if |r| < .
4
Exercise 4.3.1 (3)
lim |
1
The series is
decreases. By
P
P
an
3n − 1
2 · 5 · 8 · · · (3n − 1)
. By
=
, we see an
(−1)n+1 an = (−1)n+1
4 · 7 · 10 · · · (3n + 1)
an−1
3n + 1
√
1
an
2
1
n−1
2
=1−
<1− +o
= √
,
an−1
3n + 1
n
n
n
1
and lim √ = 0, we conclude that lim an = 0. Then by the Leibniz rule, we conclude that the
n
series converges.
Exercise 4.3.2 (1)
r )···(a+nr )
, necessary conditions for series,
Denote kn = a(a+1
b(b+1r )···(b+nr )
∞
X
(−1)n kn
n=0
to converge require kn → 0.
kn =
n
Y
a + tr
t=0
n
Y
a−b
(1 +
=
),
r
b+t
b + tr
t=0
First, we exclude the naive cases that some factors equal zero, i.e., a = −nr0 for some n0 . It
is clear under such circumstance, the series converge absolutely.(only finitely many terms are
nonzero.) Also we assume b 6= −nr for any n.
Then, kn → 0 requires: b > a if r > 0; |a + 1| < |b + 1| if r = 0; |a| < |b| or a = −b if r < 0.
a−b
1. r > 0: 0 < 1 + b+t
r < 1 finally, since we are discussing the convergence of the series,
the first finitely many terms in each kn make no contribution, we can therefore assume
a−b
without loss of generosity 0 < 1 + b+t
r < 1 for all t.
log kn =
n
X
log(1 +
t=0
a−b
)
b + tr
kn → 0 is equivalent to − log kn → ∞.
− log kn =
n
X
t=0
log(1 +
b−a
)
a + tr
b−a
b−a
−r
log(1 + a+t
as t → ∞, here the symbol ≈ means comparable. We can
r ) ≈ a+tr ≈ t
thereby deduce equivalence of kn → 0 and r ≤ 1. This is also the necessary and sufficient
condition for the series to converge conditionally.
If r = 1, − log kn ≈ log n or kn ≈ n1 , the series does not converge absolutely. If r < 1,
1−r
− log kn ≈ n1−r or kn = e−n , the series converges absolutely.
2. r = 0: It is a geometric series which converges if |a + 1| < |b + 1| both conditionally and
absolutely.
2
3. r < 0:
2b
(a) a = −b, assume without loss of generosity, b+t
r −1 > 0 for all t. kn =
Q
n
2b
2b
n
(−1)
t=0 ( b+tr − 1), we get b > 0 to make b+tr − 1 < 1, then
Qn
2b
t=0 (1− b+tr )
=
n
n
Y
X
2b
2b
log (
− 1) =
log(
− 1).
r
b+t
b + tr
t=0
t=0
We know
2b
2b
2b
−2tr
−
1)
=
log((1
+
−
2))
≈
−
2
=
≈ −tr .
b + tr
b + tr
b + tr
b + tr
Q
Qn
2b
2b
Since (−1)n kn = (−1)2n nt=0 ( b+t
r − 1) =
t=0 ( b+tr − 1), we know under this situation, the series is positive, conditional convergence is the same as absolutely convergence.
Similar to the last section of case 1, the series converges for r > −1 absolutely.
log(
r
(b) |a| < |b|, we may first assume 0 < ab < 1. Since a+t
→ ab as t → ∞, we have
b+tr
r
0 < a+t
≤ (1 − ) for t sufficiently large. Hence the series is controlled by a
b+tr
convergent geometric series, the series is convergent absolutely. The case when
−1 < ab < 0 is similar.
In summary,
Conditional convergence: 0 < r ≤ 1, b > a or r = 0, |a + 1| < |b + 1| or −1 < r < 0, a = −b
or r < 0, |a| < |b| .
Absolute convergence: 0 < r < 1, b > a or r = 0, |a + 1| < |b + 1| or −1 < r < 0, a = −b or
r < 0, |a| < |b| .
Exercise 4.3.3 (2)
We have
n sin n − (−1)n (n + 2)
n2 + sin n
n1
−
(−1)
=
= bn .
(−1)n n3 + n + 2
n
n((−1)n n3 + n + 2)
P 1
P
By comparing with
, it is easy to see that
bn absolutely converges. Therefore the
3
n
P
P
n2 + sin n
1
(absolute or conditional) convergences of
and
(−1)n are the same.
n
3
(−1) n + n + 2
n
2
P
P
1
n
+
sin
n
Since (−1)n conditionally converges, we conclude that
conditionally
n
(−1)n n3 + n + 2
converges.
Exercise 4.3.3 (5)
p
P
rn
n
The series is (−1)n an with an = p
.
By
|an | → |r|, we know the series abson (log n)q
lutely converges for |r| < 1 and diverges for |r| > 1.
For |r| = 1, by Exercise 4.2.8 (2), the series absolutely converges if and only if p > 1 or
p = 1 and q > 1.
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It remains to consider r = −1. By the Leibniz rule, the series converges for p > 0 or p = 0
and q > 0.
In summary, the series absolutely converges in the following cases: (1) |r| < 1; (2) |r| = 1,
p > 1; (2) |r| = 1, p = 1, q > 1. The series conditionally converges in the following cases: (1)
r = −1, p > 0; (2) r = −1, p = 0, q > 0.
Exercise 4.3.3 (9)
1) If a 6= 0, then a > 0. When n is large enough, an2 + bn + c > 0.
If a > 1, then
n
np+2
nn+p
n
|an |
√
·
=
=
→ 0.
1
|an2 + bn + c|
an2 + bn + c)q
an2 + bn + c
n2
the series absolutely converges.
If 0 < a < 1, then
|an | = √
n
2
an + bn + c
n
·
np
→∞
an2 + bn + c)q
so the series is diverge.
If a = 1, for any p, q, we can select k such that p + k = 2q, then
n
n
nn+p
|an |
np+k
√
=
→ 1.
=
·
1
|an2 + bn + c|
(an2 + bn + c)q
an2 + bn + c
n2
so the series absolutely converges if and only if 2q − p > 1. Note that under condition
2q − p 6 1,
n
n
|an | = √
an2 + bn + c
decreasingly goes to a constant.
When p > 2q, an doesn’t go to 0, so the series diverges. When 0 < 2q − p < 1, n is large
enough,
np+k
(an2 + bn + c)q
decreasingly goes to 0. A positive decreasing sequence multiplies the other one positive decreasing sequence will result a decreasing sequence, then an decreasingly goes to 0, then by
Dirichlet test, the series conditionally converges.
2) If a = 0, then b > 0.
If b > 1, when n is large enough
n
nn+p
n
np
√
|an | =
=
·
n
(bn + c)q
(bn + c) 2 +q
bn + c
so for any p, q, c > 0,
lim
n→∞
|an |
1
n2
4
=0
Similarily as in the case of a 6= 0, when 0 < b < 1, |an | → ∞, the series dieveges.
When b = 1, the series absolutely converges if and only if q − p > 1. If q − p 6 1,
p
1
(b + nc )n
decreasingly goes to constant √1ec . The same reason will show that when 0 < q − p 6 1, the
series converges conditionally, and when q 6 p, the series diverges.
Exercise 4.3.6
∀q > p,
X an X an
1
=
·
.
nq
np nq−p
P an
P an
Since
converges,
it
is
bounded.
So
by
Dirichlet
test,
converges.
p
n
nq
Exercise 4.3.7 (2)
This exercise is similar to the Example 4.2.7 and 4.3.5.
If 0 < a < π2 , then there exists subsequence {nk } such that kπ −
π
4
< nk a < kπ + π4 , then
| cos na|
1
1
>√
n+b
2 nk + b
whose right hand side diverges, so the original series is not absolutely convergent.
For a ∈
/ (0, π2 ), we can find b ∈ (0, π2 ) such that a + b or a − b is integer multiple of π, then
| cos na| = | cos nb|, then still we get that the series diverges.
1
decreasBut the series conditionally converges by Dirichlet test as follows. Obviously n+b
ingly goes to 0. Let’s show the partial sum
m
X
(−1)k cos na
k=1
is bounded.
Let
Sn = cos a + cos 2a + · · · + cos na.
Note that
a
sin (cos a + cos 2a + · · · + cos na
2 a
a
a
a
= sin a +
− sin a −
+ sin 2a +
− sin 2a −
+ ···
2 2 2
2
a
a
+ sin na +
− sin na −
2
2
a
a
= sin na +
− sin a −
2
2
5
that is,
Sn =
a
2
and
− sin a −
sin a1
sin na +
a
2
,
1
|Sn | 6 1 .
sin
a
Exercise 4.3.8(1)
When n is odd,
|n2
1
1
1
1
=
= 2
=
.
n
2
2
+ (−1) n |
|n − n |
n −n
n(n − 1
When n is even,
1
1
1
1
=
<
<
.
|n2 + (−1)n n2 |
n + n2
n2
n(n − 1
Therefore we always have
X
X
1
1
<
.
2
n
2
|n + (−1) n |
n(n − 1)
So the series is absolutely convergent.
Exercise 4.3.8(4)
√
Note that the series starts from n = 2, always n + (−1)n > 0
(−1) n(n−1)
2
√
√
n
n+(−1) n
=√
→1
√1
n + (−1)n
n
So the series is not absolutely converges.
Take out the odd terms, that is n = 2k + 1, we get a series
X
Ok =
X
(−1)k
√
.
2k + 1 − 1
Take out the even terms, that is n = 2k, we get a series
X
X (−1)k
√
Ek =
.
2k + 1
By Dirichlet test, both of these two series conditionally converges
For the original series, let SN be its partial sum.Then
S2k =
k
X
(Om + Em ),
m=1
6
and
S2k+1 = S2k + Ok+1
Hence the following limits exists and equal
lim S2k+1 = lim S2k
Then the original series converges.
Exercise 4.3.9(1)
We have
√
√
sin n2 + aπ = (−1)n sin( n2 + aπ − nπ) = (−1)n sin √
aπ
Since sin √
n2 + a + n
Moreover, we have
aπ
.
+a+n
√
P
is decreasing and converge to 0, the series
sin n2 + aπ converges.
n2
aπ
aπ
=
.
lim |nan | = lim n √
2
n2 + a + n P1
P
By the divergence of
. the series
|an | diverges unless a = 0. Therefore the series
n
conditionally converges.
Exercise 4.3.10
Please see the attached picture.
Exercise 4.3.12
By Dirichlet test,
X (−1)n
√
n
converges.
The new series is
∞ X
k=1
1
1
1
√
−√
−√
2k + 1
4k + 2
4k + 4
Note that
1
1
1
1
√
−√
−√
=√
2k + 1
4k + 2
4k + 4
2k + 1
and
1
1− √ −
2
r
2k + 1
4k + 1
!
r
1
2k + 1
1− √ −
4k + 1
2
is decreasing to a negative constant. So when k is large enough, the new series is a negative
series. By compare test,
q
2k+1
√ 1
√1 −
1
−
4k+1
1
1
1
2k+1
2
→ √ 1− √ −
6= 0
√1
2
2
2
k
7
so the series diverges.
Exercise 4.3.14
P (−1)n
√
Let the partial sum of
is An , and set A0 = 0
n
1. Square arrangement:
The partial sum is
S=
n
X
(ak (Ak + Ak−1 ))
k=1
n
X
=
((Ak − Ak−1 )(Ak + Ak−1 )
k=1
n
X
=
(A2k − A2k−1 )
=
k=1
A2n −
A20 = A2n → l2 ,
when l → ∞.
2. Diagonal arrangement:
The series is
X
cn
n
where
n−1
cn = (−1)
1
1
1
1
√ +p
+ ··· + p
+√
1· n
n·1
2(n − 1)
(n − 1)2
!
Since every term in the bracket is larger than n1 , |cn | > 1, so the series diverges.
8