EXERCISES FOR CHAPTER CHAPTER 5: Power Series
Find the interval of convergence of the power series below. For each state the radius of convergence.
(a)
1.
n
(b)
x
(1)
n+
n=
n= 0
1
n
x
0
Solution n+2
(1)
(a) = lim
x
n +1
n +1
(1)
n
x
=
n
,
x
1<
x < 1 converges
. At the end points,
x
=
1
the
(1)
n+
series is
n=
1
and diverges by the divergence test and at
x
=
1 the series is
0
(1)
n
n=
and diverges by the divergence test. Thus
R
=
1.
0
(b) = lim n
x
n +1
x
=
n
x
,
1 <
x < 1 converges
. At the end points,
x
=
1
(1)
n+
n=
the series is
1
and diverges by the divergence test and at
x
=
1 the series is
0
(1)
n
n=
diverges by the divergence test. Thus
(a)
2.
n=
0
x
R
=
(b)
2
0
1.
n
n +
and
n=
0
( x ) n +
n
1
Solution x
(a) = lim n
n +1
n+ x
3
=
n
n+
x
n
=
1
the series is
n=0
2
n+
3
=
x
,
1 <
x < 1 converges .
At the end points,
1 2
n +
and diverges by the integral test and at
x
=
1 the series is
n
n=
(1)
n+
2
x
lim
0 n+
2
and converges by the alternating series test. Thus ( x )
(b) = lim n
R
=
1.
n +1
2 n ( x ) n+
=
x
lim n
n +1 n+
2
=
x
,
1<
x < 1 converges .
At the end points,
n +1 x
=
1
the series is
the series is
(1)
n
n + n=0
1
n + n=0
1
1
and converges by the alternating series test and at
and diverges by the integral test. Thus
R
=
1.
x
=
1
60
K. A. Tsokos: Series and D.E.
3.
(a)
(1) n=
2n
x
n
(b)
(2 n )!
0
(2 n=
2 n +1
x
0
n +
1)!
Solution
(a) 2n+ 2
n+
(1)
x
1
(2 n + 2)!
= lim
2
= x
2n
n
(1)
2
(2 n + 2)!
n
x
n
(2 n )!
lim
= x
lim n
1 (2 n + 2)(2n + 1)
=
0 for all x and so
(2 n )!
series converges for all x . Thus
R
=
.
(b) 2n+ 3
x
= lim
(2 n + 3)! 2 n +1
2
x
n
(2 n + 1)!
lim
= x
2
= x
(2 n + 3)!
n
1
lim
(2 n + 3)(2 n + 2)
n
=
0 for all x and so series
(2 n + 1)!
converges for all x . Thus 4.
x
1 2
x
2
R
=
.
3
x
23
+
3 4
(b)
2
1 + 2 x + 3 x
+
3
4 x
+
Solution n
n+
(a) The general term is (1) n+
x
= lim
(
n n +
1)
. Hence:
1
(n + 1)(n + 2)
= x
n
x
n
x
1
(
lim n
n n +
(
n n +
1)
(n + 1)(n + 2)
. The series thus converges for
= x
1)
1 < x < 1 . At
x
=
1 , the series becomes
(1) n=
(1)
n
n+
1
(
1)
n n +
1
=
1
n=
(
1 n n +
that
1)
converges (apply for example the direct comparison test) and at
x
=
1
it is
that converges absolutely. Hence the series converges for
1
x
n
(b) The general term is
n
nx
1
. Hence, = lim n
(n + 1) x n
nx
1
= x
lim
1.
= x
(
1 n n +
Thus R
(n + 1)
n
n+
n=
(1)
=
1
1)
1.
. The series
n
converges for 1 <
x <
1.
At
x
=
1 the series becomes
n=
n
(1) that diverges by the n
1
divergence test and at
x
=
1
it is
n
n =1
converges for 1 <
x <
1.
Thus
R
=
1.
that diverges by the divergence test. Hence series
61
K. A. Tsokos: Series and D.E.
5.
(a)
5
n
( x 2)
n
2
(b)
n
1
n=
n=
( x + 3)
n
n +
1
n
1
Solution
5
n+
1
( x 2)
1 5 ( x 2)
n+
1
n +
(a) = lim
n
n
5
=
n
n
2 lim
x
n +
n
1
5
=
2 . The series converges for
x
n
1
1 < 5( x 2) < 1,
5
the series becomes series is
n =1
1
1 5
(1)
n+
n
1
1
n+
( x + 3)
2 2 ( x + 3)
9
=
1 2
< x +
(a)
the series is
n =1
1
( x 1)
and
5
R
3 lim
n +
1
n +
2
1 =
x +
1 n +
1
1 2
7
,
(1)
2
< x <
5 2
1 n +
1
that converges by the alternating series test. At
that diverges by the integral test.
7 2
x <
5 2
and
R
1 =
n
(b)
n
n=
2
.
( x 2)
1
n
n
2
n +1
( x 1)
n +1 n
( x 1)
=
x 1
n x<
the
. At the endpoints we have: at
(a)
0<
5
3 . The series thus converges for
Solution
n
=
.
5
2
=
n=
= lim
11 x
5
n
The series thus converges for
6.
3<
n=
x <
n
the series becomes
2
11
x +
5
=
1
1 < 2( x + 3) < 1,
x
5
2
=
n
n +
2
9 x
1
n +
n
n
5
. At the endpoints we have: at
that diverges being the harmonic series.
(b) = lim
=
5
11
n
2
x
< x <
that converges by the alternating series test. At
The series thus converges for
7
9
2< ,
n
n=
< x
2 converges
lim n
n n +1
=
x 1,
1<
x 1 < 1,
62
K. A. Tsokos: Series and D.E.
For
x
=
series becomes
0
1
For
x
=
n
converges by the alternating series test.
n
1
n diverges being a p=1 series.
series becomes
2
(1)
1
Series converges for (b) n 1 ( x 2)
0
x <
2.
Thus
R
=
1.
+
2
2
(n + 1)
= lim n
( x 2) n
1<
x <
=
n
n x 2 lim n + 1
x
=
n
series becomes
1
(1) n
x
=
3
series becomes
1
Series converges for
7.
(a)
2 , 1 < x 2 < 1,
3 converges
1
For
x
2
For
=
n
1
x
1 n
3.
that converges absolutely.
2
that converges being a p =2 series.
2
Thus
R
=
1.
n
x
(b)
n
1
n =1
n
x
n +
3
Solution
(a) x
n +1
n +1
= lim
x
n
=
n
n
lim
x
n +1
n
=
x
,
1<
x < 1 converges
. At the endpoints the series
n
n
becomes
(1)
that converges by the alternating series test and
n
1
1
harmonic series and diverges. Hence the series converges for
1
x <
1
that is the
n
1.
Thus
R
1.
=
(b) x
= lim
n +1
n+ x
n
4
=
n
n+
x
lim n
n+
3
n+
4
=
x
,
1<
x < 1 converges
3
series becomes
1
(1)
n
n +
3
that converges by the alternating series test and
8.
(a)
n= 2
ln n n
n
x
1
diverges by the integral test. Hence the series converges for
. At the endpoints the
(b)
n= 2
1
ln n n
2
x <
n
x
1.
1 n +
Thus
R
3 =
that 1.
63
K. A. Tsokos: Series and D.E.
Solution
(a) ln(n + 1) x
n +1
n +1
= lim
ln n
n
x
=
n
lim
x
n
n
ln(n + 1)
n +1
ln n
=
x
,
1<
x < 1 converges
. At the
n
endpoints,
x
1 the series is
=
n=
and at
x
the series is
1
=
n=2
Hence series converges for (b) ln(n + 1) x ln n n
x
ln n
and diverges by the direct comparison test with
n
x
1
=
x <
n
n
lim
x
=
the series is
1
n=
converges for
9.
R
=
1
1.
2
ln(n + 1) ln n
=
x
1
x
1.
ln n
2 n
2
Thus
ln n
2 n
2
,
1<
x < 1 converges
. At the
n
(1) and converges by the alternating series test
and converges by the integral test. Hence series
R
=
1.
(a) 4
n
x
(b)
(1) 1
1
!
1 n
n+
n
2
n
x
Solution
(a) n +1
x 4 x 4
= lim n
=
n
x
4
4 < x < 4 converges . At the endpoints the series is
,
( 1)
n
and
1 , both diverge. Hence series converges for
1
4 < x < 4 . Thus
R
=
lim (n + 1) = unless
x
=
4.
1
(b) n+
n+
(1)
2
(n + 1)! x
n n+
(1)
!
=
0.
n
1 n x 2 n
R
1
2
(n + 1)
= lim
Thus
.
n
2
x
Thus
2
n=
and at
1.
(n + 1)
n
1 the series is
=
n =2
endpoints,
n
n
2
2
n
(1) and converges by the alternating series test
n +1
(n + 1)
= lim
ln n
= x
lim n
n
2
(n + 1)! 2
(n + 1)
n
!
= x
n
0
.
64
K. A. Tsokos: Series and D.E.
10.
(a)
(1) 2
1
n=
n
n
( x 2)
2n
n
(b)
(2 x 1)
n
n
1
Solution
(a) (n + 1) = lim n
n
2<
( x 2)
n +1
2n+2
2 n ( x 2) 2
4
=
n4
=
2
=
2
,
4
4
<
x
2 < 4,
n diverges by the divergence test.
1
x
x
n
2n
2 series becomes
For
n +1
n
=
6 converges
x<
x
n
lim
2n
For
2
x
=
6
series becomes
1
n
(1) n 4 2
1
n =
2n
(1) n n diverges by the alternating series 1
test. Series converges for 2 <
6.
x <
Thus
R
=
4.
(b) n +1
(2 x 1) n +1
= lim
n
(2 x 1)
n
=
2 x 1 lim n
n n +1
2 x 1 ,
=
1 < 2 x 1 < 1,
n
0<
x < 1 converges
For
x
=
0
series becomes
1
For
x
=
1
series becomes
(1)
n
converges by the alternating series test.
n
1
n diverges being a p series with p=1. 1
Series converges for
0
. Thus
x < 1
R
1 =
2
.
11.
(a)
2
n
n
(b)
x
1
n
x
2
n
1
Solution
(a) = lim n
For
x
=±
1 2
2
n +1
x
2
n
x
n +1
n
=
2
x
,
1<
2 x
< 1,
1
2
<
x<
1 2
converges .
series becomes
1 diverges by divergence test and 1
(1) n diverges by divergence test. 1
Series converges for
1 2
< x <
1 2
. Thus
R
1 =
2
.
65
K. A. Tsokos: Series and D.E.
x
(b) = lim 2
n +1 n +1
x
n
2
x
=
n
2
,
1 <
x
< 1,
2
2
<
2 converges .
x <
n
For
x
series becomes
= ±2
1 diverges by divergence test and (1) n diverges 1
1
by divergence test. Series converges for 2 <
x <
2.
Thus
R
2.
=
12.
(a)
n
2
( x 2)
n
(b)
1
n
n
n x
1
Solution
(a) 2
= lim
(n + 1) ( x 2)
n
1<
x<
n
2
( x 2)
n+2
2
=
n
2 lim
x
(n + 1)
n
n
2
=
x
2,
1<
2 <1
x
3 converges
At the endpoints we have:
x
=
and the series becomes
1
n
2
(1) that diverges by the n
1
divergence test. At
x
=
3
and the series becomes
n
2
that diverges by the divergence
1
1<
test. Hence the series converges for
x <
3.
Thus
R
=
1.
(b) n+
= lim
(n + 1) n
=
0
. Thus
13.
(a)
n+
x
1
= x
n
n x
n
x
1
!
=
n +
n
n
n+
1
1
n+
n =
x
1 lim 1 + n
n
1 n =
x e
lim n = unless n
0.
n
n x
(2 1
R
lim
(b)
)!
n
1
4
n
n x n!
Solution
(a) n+
(n + 1)! x = lim n
(2 n + 2)! !
n
n x
(2 n)!
(b)
1
= x
lim n
n +
1
(2 n + 2)(2 n + 1)
=
0 for all values of x . Thus
R
=
.
66
K. A. Tsokos: Series and D.E. 4
(n + 1)
n+
x
(n + 1)!
= lim
4
n
(1)
+ 1)
1
lim n
(n
n +
4
1
n
=
0 for all values of x . Thus
R
=
.
!
(a)
= x
n
n x
n
14.
1
1
n+
(2 x 1) n +
1
n
(b)
1
(2
n
+
n
n
3 ) x
1
Solution
(a) n +1
(2 x 1) n +1
= lim
2 x 1 ,
=
(2 x 1)
n
n
1 < 2 x 1 < 1,
0
<
x < 1 converges
n
For
x
=
0
series becomes 1
For
x
=
1
series becomes
0
< x
n +1
(1)
diverges by the integral test.
n +1
converges by the alternating series test.
n +1
1
Series converges for
1
1.
Thus
R
1 =
2
.
(b) n +1
= lim n
(2
n +1
(2
+
n
+
1 < 3 x < 1, For
x
=±
1 3
n +1
3
n
) x
3 ) x 1 3
<
n +1
=
n
lim
x
n
x<
1 3
2 ( + 1) 3 1 2 1 3 ( + ) 3 3 3 3 n
n
=
3
x
converges
series becomes
2 ( 3 1
n
2 + 1) and ( 3 1
n
1)
+ 1)(
n
and both diverge by
divergence test and alternating series test. Series converges for
1 3
1 < x <
3
. Thus
R
1 =
3
.
15.
The Bessel function
J 0 ( x )
may be defined by the power series
(1)
k
k
=
0
Find the radius of convergence of this series. (b) The Bessel function defined through
J 1 ( x )
d =
dx
J 0 ( x ) .
Find the power series for
J 1 ( x ) .
x
2 k
2 k
2
2 ( k !) J 1 ( x )
. (a)
may be
67
K. A. Tsokos: Series and D.E.
Solution
(a) 2n+2
n+
(1)
x
1
2
= lim
2n+2
2n
n
(1) =
2
2
2n
2
x
=
2
(n!)
lim
4
x
n
R
((n + 1)!)
x
((n + 1)!)
n
=
2
lim
4
n
1 2
(n + 1)
=
0 for all x . Hence
2
2 (n!)
.
(b)
J 1 ( x ) =
(1)
2 k
d
(1)
(1) k = 0
2 k
(2 k ) x
(1)
2
(1)
2 k 1
x
k
2
k = 0
2
x
=
2 k 1
2 k
k + 1
k = 1
2
2 ( k !)
=
2 k
2 ( k !)
k = 1
k
2
x
k
dx
2 k
2 ( k !)
k = 0
=
x
k
dx
=
d
2 k + 1
2 k 1
( k 1)! k !
2 k + 1
( k + 1)! k !
16.
Find the sum of the power series
2 n 1
nx
by using another power series of known
n =1
sum. Solution
Since,
2n
x
=
2
1+
x
4
+ x
1 +
…=
n=0
d
x dx n
=
2n
=
0
2nx n
=
nx
17.
2 n 1
(for 1 <
x <
2 n 1
=
0
n
2
1 x
2nx n
1
=
2 n 1
1)
d =
we have by differentiation that
2 x
1
dx 1 x
2
=
2 2
(1 x )
x =
1
=
2 2
(1 x )
dy
xy with initial condition y 1 when x 0 . dx (a) This is a separable differential equation. Solve this equation. (b) Now try to solve the equation again by assuming that the solution can be written as a power
Consider the differential equation
=
=
series,
2
y = a0 + a1 x + a2 x + =
a x n
n=0
n
. Calculate
dy dx
n
and then
xy
=
and equate the
coefficients of x on both sides to find the value of a . Hence find the solution of the differential equation as a power series. (c) Find the interval of convergence of the power series you got in (b). n
68
K. A. Tsokos: Series and D.E. Solution
(a)
dy dx
=
dy
xy
y
x
C
=
1
and so
y
=
xdx ln y =
=
e
x 2
x 2
ln C y
=
Ce 2 . The initial condition means that
a1 x + a2 x 2
+
+
2
2
2
.
(b) y = a0
+
yx = a0 x + a1 x 2
dy dx
=
+
a2 x
a1 + 2 a2 x + 3a3 x 2
3
+
+
and so a
1
+
2
2 a2 x + 3a3 x
+
= a0 x + a1 x 2 + a2 x 3 +
Now, the initial condition dictates that a0 1 . Since there is no constant term on the right side we must have a1 0 . Matching coefficients of equal powers of x we have that =
=
2 a2
=
a
3a3
=
a
4 a4
0
1
1
=
0
=
1 a
2
a3
=
=
2 0
1 a
=
2
=
2
1 a4
1
=
2
4
and in general a
2 n +1
=
0
1 a
2n
=
2n
so that the solution is
y
=
x
2 n
=
1
0
2n 2
1
1
2n 4
2
1 =
2
n
2
2n
n
n!
x
n!
. We may then deduce that
e
2
=
2 n =0
2n
x
n
n!
.
(c) Applying the power series ratio test gives 2n+2
x
= lim n
2
n+
1
(n + 1)! 2n
x
2
n
n
!
=
1
2
x
2
lim n
n
!
(n + 1)!
=
1
2
x
2
lim n
1 n +
1
=
0 , for all values of x .
