Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 6 doc

12.8 Hints
Hint 12.1
Use the Cauchy convergence criterion for series. In particular, consid er |S
N+1
− S
N
|.
Hint 12.2
CONTINUE
Hint 12.3
1.
∞

n=2
1
n ln(n)
Use the integral test.
2.
∞

n=2
1
ln (n
n
)
Simplify the summand.
3.
∞

n=2

ln
n
√
ln n
Simplify the summand. Use the comparison test.
4.
∞

n=10
1
n(ln n)(ln(ln n))
Use the integral test.
574
5.
∞

n=1
ln (2
n
)
ln (3
n
) + 1
Show that the terms in the sum do not vanish as n → ∞
6.
∞

n=0
1
ln(n + 20)

Shift the indices.
7.
∞

n=0
4
n
+ 1
3
n
− 2
Show that the terms in the sum do not vanish as n → ∞
8.
∞

n=0
(Log
π
2)
n
This is a geometric series.
9.
∞

n=2
n
2
− 1
n
4

− 1
Simplify the integrand. Use the comparison test.
10.
∞

n=2
n
2
(ln n)
n
Compare to a geometric series.
575
11.
∞

n=2
(−1)
n
ln

1
n

Group pairs of consecutive terms to obtain a series of positive terms.
12.
∞

n=2
(n!)
2

(2n)!
Use the comparison test.
13.
∞

n=2
3
n
+ 4
n
+ 5
5
n
− 4
n
− 3
Use the root test.
14.
∞

n=2
n!
(ln n)
n
Show that the terms do not vanish as n → ∞.
15.
∞

n=2
e

n
ln(n!)
Show that the terms do not vanish as n → ∞.
16.
∞

n=1
(n!)
2
(n
2
)!
Apply the ratio test.
576
17.
∞

n=1
n
8
+ 4n
4
+ 8
3n
9
− n
5
+ 9n
Use the comparison test.
18.

∞

n=1

1
n
−
1
n + 1

Use the comparison test.
19.
∞

n=1
cos(nπ)
n
Simplify the integrand.
20.
∞

n=2
ln n
n
11/10
Use the integral test.
Hint 12.4
Group the terms.
1 −
1

2
=
1
2
1
3
−
1
4
=
1
12
1
5
−
1
6
=
1
30
···
577
Hint 12.5
Show that
|S
2n
− S
n
| >
1

2
.
Hint 12.6
The alternating harmonic series is conditionally convergent. Let {a
n
} and {b
n
} be the positive and negative terms in
the sum, respectively, ordered in decreasing magnitude. Note that both

∞
n=1
a
n
and

∞
n=1
b
n
are divergent. Devise a
method for alternately taking terms from {a
n
} and {b
n
}.
Hint 12.7
Use the ratio test.
Hint 12.8
Use the integral test.

Hint 12.9
Note that sin(nx) = (
e
ınx
). This substitute will yield a finite geometric series.
Hint 12.10
Let S
n
be the sum. Consider S
n
− zS
n
. Use the finite geometric sum.
Hint 12.11
1. The summand is a rational function. Find the first few partial sums.
2.
3. This a geometric series.
Hint 12.12
CONTINUE
578
Hint 12.13
CONTINUE
1.
∞

n=0
z
n
(z + 3)
n

2.
∞

n=2
Log z
ln n
3.
∞

n=1
z
n
4.
∞

n=1
(z + 2)
2
n
2
5.
∞

n=1
(z −
e
)
n
n
n

6.
∞

n=1
z
2n
2
nz
7.
∞

n=0
z
n!
(n!)
2
8.
∞

n=0
z
ln(n!)
n!
9.
∞

n=0
(z −π)
2n+1
n

π
n!
579
10.
∞

n=0
ln n
z
n
Hint 12.14
Hint 12.15
CONTINUE
Hint 12.16
Differentiate the geometric series. Integrate the geometric series.
Hint 12.17
The Taylor series is a geometric series.
Hint 12.18
Hint 12.19
Hint 12.20
1.
1
z
=
1
1 + (z −1)
The right side is the sum of a geometric series.
2. Integrate the series for 1/z.
3. Differentiate the series for 1/z.
4. Integrate the series for Log z.

580
Hint 12.21
Evaluate the derivatives of
e
z
at z = 0. Use Taylor’s Theorem.
Write the cosine and sine in terms of the exponential function.
Hint 12.22
cos z = −cos(z − π)
sin z = −sin(z − π)
Hint 12.23
CONTINUE
Hint 12.24
CONTINUE
Hint 12.25
Hint 12.26
Hint 12.27
Hint 12.28
Hint 12.29
Hint 12.30
CONTINUE
581
12.9 Solutions
Solution 12.1

∞
n=0
a
n
converges only if the partial sums, S

n
, are a Cauchy sequence.
∀ > 0 ∃N s.t. m, n > N ⇒ |S
m
− S
n
| < ,
In particular, we can consider m = n + 1.
∀ > 0 ∃N s.t. n > N ⇒ |S
n+1
− S
n
| < 
Now we note that S
n+1
− s
n
= a
n
.
∀ > 0 ∃N s.t. n > N ⇒ |a
n
| < 
This is exactly the Cauchy convergence criterion for the sequence {a
n
}. Thus we see that lim
n→∞
a
n
= 0 is a necessary

condition for the convergence of the series

∞
n=0
a
n
.
Solution 12.2
CONTINUE
Solution 12.3
1.
∞

n=2
1
n ln(n)
Since this is a series of positive, monotone decreasing terms, the sum converges or diverges with the integral,

∞
2
1
x ln x
dx =

∞
ln 2
1
ξ
dξ
Since the integral diverges, the series also diverges.

2.
∞

n=2
1
ln (n
n
)
=
∞

n=2
1
n ln(n)
582
The sum converges.
3.
∞

n=2
ln
n
√
ln n =
∞

n=2
1
n
ln(ln n) ≥

∞

n=2
1
n
The sum is divergent by the comparison test.
4.
∞

n=10
1
n(ln n)(ln(ln n))
Since this is a series of positive, monotone decreasing terms, the sum converges or diverges with the integral,

∞
10
1
x ln x ln(ln x)
dx =

∞
ln(10)
1
y ln y
dy =

∞
ln(ln(10))
1
z

dz
Since the integral diverges, the series also diverges.
5.
∞

n=1
ln (2
n
)
ln (3
n
) + 1
=
∞

n=1
n ln 2
n ln 3 + 1
=
∞

n=1
ln 2
ln 3 + 1/n
Since the terms in the sum do not vanish as n → ∞, the series is divergent.
6.
∞

n=0
1

ln(n + 20)
=
∞

n=20
1
ln n
The series diverges.
7.
∞

n=0
4
n
+ 1
3
n
− 2
Since the terms in the sum do not vanish as n → ∞, the series is divergent.
583
8.
∞

n=0
(Log
π
2)
n
This is a geometric series. Since |Log
π

2| < 1, the series converges.
9.
∞

n=2
n
2
− 1
n
4
− 1
=
∞

n=2
1
n
2
+ 1
<
∞

n=2
1
n
2
The series converges by comparison to the harmonic series.
10.
∞


n=2
n
2
(ln n)
n
=
∞

n=2

n
2/n
ln n

n
Since n
2/n
→ 1 as n → ∞, n
2/n
/ ln n → 0 as n → ∞. The series converges by comparison to a geometric
series.
11. We group pairs of consecutive terms to obtain a series of positive terms.
∞

n=2
(−1)
n
ln

1

n

=
∞

n=1

ln

1
2n

− ln

1
2n + 1

=
∞

n=1
ln

2n + 1
2n

The series on the right side diverges because the terms d o not vanish as n → ∞.
12.
∞


n=2
(n!)
2
(2n)!
=
∞

n=2
(1)(2) ···n
(n + 1)(n + 2) ···(2n)
<
∞

n=2
1
2
n
The series converges by comparison with a geometric series.
584
13.
∞

n=2
3
n
+ 4
n
+ 5
5
n

− 4
n
− 3
We u se the root test to check for convergence.
lim
n→∞
|a
n
|
1/n
= lim
n→∞




3
n
+ 4
n
+ 5
5
n
− 4
n
− 3





1/n
= lim
n→∞
4
5




(3/4)
n
+ 1 + 5/4
n
1 − (4/5)
n
− 3/5
n




1/n
=
4
5
< 1
We s ee that the series is absolutely convergent.
14. We will use the comparison test.
∞


n=2
n!
(ln n)
n
>
∞

n=2
(n/2)
n/2
(ln n)
n
=
∞

n=2


n/2
ln n

n
Since the terms in the series on the right side do not vanish as n → ∞, the series i s divergent.
15. We will use the comparison test.
∞

n=2
e
n
ln(n!)

>
∞

n=2
e
n
ln(n
n
)
=
∞

n=2
e
n
n ln(n)
Since the terms in the series on the right side do not vanish as n → ∞, the series i s divergent.
585
16.
∞

n=1
(n!)
2
(n
2
)!
We app ly the ratio test.
lim
n→∞





a
n+1
a
n




= lim
n→∞




((n + 1)!)
2
(n
2
)!
((n + 1)
2
)!(n!)
2





= lim
n→∞




(n + 1)
2
((n + 1)
2
− n
2
)!




= lim
n→∞




(n + 1)
2
(2n + 1)!





= 0
The series is convergent.
17.
∞

n=1
n
8
+ 4n
4
+ 8
3n
9
− n
5
+ 9n
=
∞

n=1
1
n
1 + 4n
−4
+ 8n
−8
3 − n
−4
+ 9n

−8
>
1
4
∞

n=1
1
n
We s ee that the series is divergent by comparison to the harmonic series.
18.
∞

n=1

1
n
−
1
n + 1

=
∞

n=1
1
n
2
+ n
<

∞

n=1
1
n
2
The series converges by the comparison test.
586
19.
∞

n=1
cos(nπ)
n
=
∞

n=1
(−1)
n
n
We recognize this as the alternating harmonic series, which is conditionally convergent.
20.
∞

n=2
ln n
n
11/10
Since this is a series of positive, monotone decreasing terms, the sum converges or diverges with the integral,


∞
2
ln x
x
11/10
dx =

∞
ln 2
y
e
−y/10
dy
Since the integral is convergent, so is the series.
Solution 12.4
∞

n=1
(−1)
n+1
n
=
∞

n=1

1
2n − 1
−

1
2n

=
∞

n=1
1
(2n − 1)(2n)
<
∞

n=1
1
(2n − 1)
2
<
1
2
∞

n=1
1
n
2
=
π
2
12
Thus the series is convergent.

587
Solution 12.5
Since
|S
2n
− S
n
| =





2n−1

j=n
1
j





≥
2n−1

j=n
1
2n − 1
=

n
2n − 1
>
1
2
the series does not satisfy the Cauchy convergence criterion.
Solution 12.6
The alternating harmonic series is conditionally convergent. That is, the sum is convergent but not absolutely conver-
gent. Let {a
n
} and {b
n
} be the positive and negative terms in the sum, respectively, ordered in decreasing magnitude.
Note that both

∞
n=1
a
n
and

∞
n=1
b
n
are divergent. Otherwise the alternating harmonic series would be absolutely
convergent.
To sum the terms in the series to π we repeat the following two steps indefinitely:
1. Take terms from {a
n

} until the sum is greater than π.
2. Take terms from {b
n
} until the sum is less than π.
Each of these steps can always be accomplished be cause the sums,

∞
n=1
a
n
and

∞
n=1
b
n
are both divergent. Hence the
tails of the series are divergent. No matter how many terms we take, the remaining terms in each series are divergent.
In each step a finite, nonzero number of terms f rom the respective series is taken. Thus all the terms will be used.
Since the terms in each series vanish as n → ∞, the running sum converges to π.
588
Solution 12.7
Applying the ratio test,
lim
n→∞




a

n+1
a
n




= lim
n→∞
(n + 1)!n
n
n!(n + 1)
(n+1)
= lim
n→∞
n
n
(n + 1)
n
= lim
n→∞

n
(n + 1)

n
=
1
e
< 1,

we see that the series is absolutely convergent.
Solution 12.8
The harmonic series,
∞

n=1
1
n
α
= 1 +
1
2
α
+
1
3
α
+ ··· ,
converges or diverges absolutely with the integral,

∞
1
1
|x
α
|
dx =

∞
1

1
x
(α)
dx =

[ln x]
∞
1
for (α) = 1,

x
1−(α)
1−(α)

∞
1
for (α) = 1.
The integral converges only for (α) > 1. Thus the harmonic series converges absolutely for (α) > 1 and diverges
absolutely for (α) ≤ 1.
589
Solution 12.9
N−1

n=1
sin(nx) =
N−1

n=0
sin(nx)
=

N−1

n=0
(
e
ınx
)
= 

N−1

n=0
(
e
ıx
)
n

=

(N) for x = 2πk


1−
e
ınx
1−
e
ıx


for x = 2πk
=

0 for x = 2πk


e
−ıx/2
−
e
ı(N−1/2)x
e
−ıx/2
−
e
ıx/2

for x = 2πk
=

0 for x = 2πk


e
−ıx/2
−
e
ı(N−1/2)x
−ı2 sin(x/2)


for x = 2πk
=

0 for x = 2πk


e
−ıx/2
−
e
ı(N−1/2)x
2 sin(x/2)

for x = 2πk
N−1

n=1
sin(nx) =

0 for x = 2πk
cos(x/2)−cos((N−1/2)x)
2 sin(x/2)
for x = 2πk
590
Solution 12.10
Let
S
n
=
n


k=1
kz
k
.
S
n
− zS
n
=
n

k=1
kz
k
−
n

k=1
kz
k+1
=
n

k=1
kz
k
−
n+1


k=2
(k − 1)z
k
=
n

k=1
z
k
− nz
n+1
=
z −z
n+1
1 − z
− nz
n+1
n

k=1
kz
k
=
z(1 − (n + 1)z
n
+ nz
n+1
)
(1 − z)
2

Let
S
n
=
n

k=1
k
2
z
k
.
S
n
− zS
n
=
n

k=1
(k
2
− (k − 1)
2
)z
k
− n
2
z
n+1

= 2
n

k=1
kz
k
−
n

k=1
z
k
− n
2
z
n+1
= 2
z(1 − (n + 1)z
n
+ nz
n+1
)
(1 − z)
2
−
z −z
n+1
1 − z
− n
2

z
n+1
591
n

k=1
k
2
z
k
=
z(1 + z −z
n
(1 + z + n(n(z −1) −2)(z − 1)))
(1 − z)
3
Solution 12.11
1.
∞

n=1
a
n
=
1
2
+
1
6
+

1
12
+
1
20
+ ···
We conjecture that the terms in the sum are rational functions of summation index. That is, a
n
= 1/p(n) where
p(n) is a polynomial. We use divided differences to determine the order of the polynomial.
2 6 12 20
4 6 8
2 2
We s ee that the polynomial is second order. p(n) = an
2
+ bn + c. We solve for the coefficients.
a + b + c = 2
4a + 2b + c = 6
9a + 3b + c = 12
p(n) = n
2
+ n
We e xamine the first few partial sums.
S
1
=
1
2
S
2

=
2
3
S
3
=
3
4
S
4
=
4
5
592
We conjecture that S
n
= n/(n+1). We prove this with induction. The base case is n = 1. S
1
= 1/(1+1) = 1/2.
Now we assume the ind uction hypothesis and calculate S
n+1
.
S
n+1
= S
n
+ a
n+1
=
n

n + 1
+
1
(n + 1)
2
+ (n + 1)
=
n + 1
n + 2
This proves the induction hypothesis. We calculate the limit of the partial sums to evaluate the series.
∞

n=1
1
n
2
+ n
= lim
n→∞
n
n + 1
∞

n=1
1
n
2
+ n
= 1
2.

∞

n=0
(−1)
n
= 1 + (−1) + 1 + (−1) + ···
Since the terms in the series do not vanish as n → ∞, the series is divergent.
3. We can directly sum this geometric series.
∞

n=1
1
2
n−1
1
3
n
1
5
n+1
=
1
75
1
1 − 1/30
=
2
145
CONTINUE
593

Solution 12.12
The innermost sum is a geometric series.
∞

k
n
=k
n−1
1
2
k
n
=
1
2
k
n−1
1
1 − 1/2
= 2
1−k
n−1
This gives us a relationship between n nested sums and n − 1 nested sums.
∞

k
1
=0
∞


k
2
=k
1
···
∞

k
n
=k
n−1
1
2
k
n
= 2
∞

k
1
=0
∞

k
2
=k
1
···
∞


k
n−1
=k
n−2
1
2
k
n−1
We e valuate the n nested sums by induction.
∞

k
1
=0
∞

k
2
=k
1
···
∞

k
n
=k
n−1
1
2
k

n
= 2
n
Solution 12.13
CONTINUE.
1.
∞

n=0
z
n
(z + 3)
n
2.
∞

n=2
Log z
ln n
3.
∞

n=1
z
n
4.
∞

n=1
(z + 2)

2
n
2
594
5.
∞

n=1
(z −
e
)
n
n
n
6.
∞

n=1
z
2n
2
nz
7.
∞

n=0
z
n!
(n!)
2

8.
∞

n=0
z
ln(n!)
n!
9.
∞

n=0
(z −π)
2n+1
n
π
n!
10.
∞

n=0
ln n
z
n
Solution 12.14
1. We assume that β = 0. We determine the radius of convergence with the ratio test.
R = lim
n→∞





a
n
a
n+1




= lim
n→∞




(α − β) ···(α −(n −1)β)/n!
(α − β) ···(α −nβ)/(n + 1)!




= lim
n→∞




n + 1
α − nβ





=
1
|β|
595
The series converges absolutely for |z| < 1/|β|.
2. By the ratio test formula, the radius of absolute convergence is
R = lim
n→∞




n/2
n
(n + 1)/2
n+1




= 2 lim
n→∞




n

n + 1




= 2
By the root test formula, the radius of absolute convergence is
R =
1
lim
n→∞
n

|n/2
n
|
=
2
lim
n→∞
n
√
n
= 2
The series converges absolutely for |z −ı| < 2.
3. We determine the radius of convergence with the Cauchy-Hadamard formula.
R =
1
lim sup
n


|a
n
|
=
1
lim sup
n

|n
n
|
=
1
lim sup n
= 0
The series converges only for z = 0.
596
4. By the ratio test formula, the radius of absolute convergence is
R = lim
n→∞




n!/n
n
(n + 1)!/(n + 1)
n+1





= lim
n→∞




(n + 1)
n
n
n




= lim
n→∞

n + 1
n

n
= exp

lim
n→∞
ln


n + 1
n

n

= exp

lim
n→∞
n ln

n + 1
n

= exp

lim
n→∞
ln(n + 1) − ln(n)
1/n

= exp

lim
n→∞
1/(n + 1) − 1/n
−1/n
2

= exp


lim
n→∞
n
n + 1

=
e
1
The series converges absolutely in the circle, |z| <
e
.
5. By the Cauchy-Hadamard formula, the radius of absolute convergence is
R =
1
lim sup
n

|(3 + (−1)
n
)
n
|
=
1
lim sup (3 + (−1)
n
)
=
1

4
597