power series

1. A power series P(x) is a "polynomial" in x of infinitely
many terms.
Power Series

2. A power series P(x) is a "polynomial" in x of infinitely
many terms. That is,
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 + a4x4 ….
Power Series
P(x) =

3. A power series P(x) is a "polynomial" in x of infinitely
many terms. That is,
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 + a4x4 ….
Note that P(0) = a0.
Power Series
P(x) =

4. A power series P(x) is a "polynomial" in x of infinitely
many terms. That is,
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 + a4x4 ….
The basic question for a power series is to determine
the value(s) of x where it converges or diverges.
Note that P(0) = a0.
Power Series
P(x) =

5. A power series P(x) is a "polynomial" in x of infinitely
many terms. That is,
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 + a4x4 ….
Example: A. Let P(x) = 1 + x + x2 + x3 + x4 …
The basic question for a power series is to determine
the value(s) of x where it converges or diverges.
Note that P(0) = a0.
Power Series
P(x) =

6. A power series P(x) is a "polynomial" in x of infinitely
many terms. That is,
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 + a4x4 ….
Example: A. Let P(x) = 1 + x + x2 + x3 + x4 …
From the theorem of geometric series, given x = r,
I. P(r) = 1 + r + r2 + r3 + r4 + … converges if | r | < 1.
The basic question for a power series is to determine
the value(s) of x where it converges or diverges.
Note that P(0) = a0.
Power Series
P(x) =

7. A power series P(x) is a "polynomial" in x of infinitely
many terms. That is,
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 + a4x4 ….
Example: A. Let P(x) = 1 + x + x2 + x3 + x4 …
From the theorem of geometric series, given x = r,
I. P(r) = 1 + r + r2 + r3 + r4 + … converges if | r | < 1.
II. P(r) diverges if | r | > 1.
The basic question for a power series is to determine
the value(s) of x where it converges or diverges.
Note that P(0) = a0.
Power Series
P(x) =

8. A power series P(x) is a "polynomial" in x of infinitely
many terms. That is,
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 + a4x4 ….
Example: A. Let P(x) = 1 + x + x2 + x3 + x4 …
From the theorem of geometric series, given x = r,
I. P(r) = 1 + r + r2 + r3 + r4 + … converges if | r | < 1.
II. P(r) diverges if | r | ≥ 1.
The basic question for a power series is to determine
the value(s) of x where it converges or diverges.
Note that P(0) = a0.
Power Series
P(x) =
0
–1 converges (absolutely) 1 divergesdiverges

9. Example: B. Let P(x) = 1 +
Power Series
x x2
+
2
+ x3
3
+ x4
4
+..

10. Example: B. Let P(x) = 1 +
Use the ceiling theorem:
Power Series
x x2
+
2
+ x3
3
+ x4
4
+..
P(x) = 1 +|x|
|x|2
+
2
+
3
+
4
+..
|x|3 |x|4
1 +|x| + |x2| + |x3| + |x4| +.. …
vI vI vI vI vI

11. Example: B. Let P(x) = 1 +
Use the ceiling theorem:
Power Series
x x2
+
2
+ x3
3
+ x4
4
+..
P(x) = 1 +|x|
|x|2
+
2
+
3
+
4
+..
|x|3 |x|4
1 +|x| + |x2| + |x3| + |x4| +.. …
vI vI vI vI vI
and 1 +|x| + |x2| + |x3| + |x4| +… converges for | x | < 1,
we conclude that

12. Example: B. Let P(x) = 1 +
Use the ceiling theorem:
Power Series
x x2
+
2
+ x3
3
+ x4
4
+..
P(x) = 1 +|x|
|x|2
+
2
+
3
+
4
+..
|x|3 |x|4
1 +|x| + |x2| + |x3| + |x4| +.. …
vI vI vI vI vI
and 1 +|x| + |x2| + |x3| + |x4| +… converges for | x | < 1,
we conclude that
P(x) = 1 +|x|
|x|2
+
2
+
3
+
4
+..
|x|3 |x|4
also converges absolutely for |x| < 1.

13. Power Series
For x = ±1:

14. P(1) = 1 + is basically the
harmonic series, so it diverges.
Power Series
1 1+
2
+ 1
3
+ 1
4
+..
For x = ±1:

15. P(1) = 1 + is basically the
harmonic series, so it diverges.
Power Series
1 1+
2
+ 1
3
+ 1
4
+..
P(–1) = 1 – is basically the
alternating harmonic series. It converges conditionally.
1 1 –
2
+ 1
3
+ 1
4 – ..
For x = ±1:

16. P(1) = 1 + is basically the
harmonic series, so it diverges.
Power Series
1 1+
2
+ 1
3
+ 1
4
+..
For x = | r | > 1,
P(–1) = 1 – is basically the
alternating harmonic series. It converges conditionally.
1 1 –
2
+ 1
3
+ 1
4 – ..
For x = ±1:

17. P(1) = 1 + is basically the
harmonic series, so it diverges.
Power Series
1 1+
2
+ 1
3
+ 1
4
+..
For x = | r | > 1, the general terms of P(r)
P(–1) = 1 – is basically the
alternating harmonic series. It converges conditionally.
1 1 –
2
+ 1
3
+ 1
4 – ..
For x = ±1:
rn
n ∞ as n  ∞

18. P(1) = 1 + is basically the
harmonic series, so it diverges.
Power Series
1 1+
2
+ 1
3
+ 1
4
+..
For x = | r | > 1, the general terms of P(r)
P(–1) = 1 – is basically the
alternating harmonic series. It converges conditionally.
1 1 –
2
+ 1
3
+ 1
4 – ..
For x = ±1:
rn
n ∞ as n  ∞ (verify this by L'Hopital's rule).

19. P(1) = 1 + is basically the
harmonic series, so it diverges.
Power Series
1 1+
2
+ 1
3
+ 1
4
+..
For x = | r | > 1, the general terms of P(r)
P(r) = 1 + r r2
+
2
+
3
+
4
+.. diverges.r3 r4
P(–1) = 1 – is basically the
alternating harmonic series. It converges conditionally.
1 1 –
2
+ 1
3
+ 1
4 – ..
For x = ±1:
rn
n ∞ as n  ∞ (verify this by L'Hopital's rule).
Hence

20. P(1) = 1 + is basically the
harmonic series, so it diverges.
Power Series
1 1+
2
+ 1
3
+ 1
4
+..
For x = | r | > 1, the general terms of P(r)
P(r) = 1 + r r2
+
2
+
3
+
4
+.. diverges.r3 r4
P(–1) = 1 – is basically the
alternating harmonic series. It converges conditionally.
1 1 –
2
+ 1
3
+ 1
4 – ..
For x = ±1:
rn
n ∞ as n  ∞ (verify this by L'Hopital's rule).
Hence
We summarize the result using the following picture:

21. P(1) = 1 + is basically the
harmonic series, so it diverges.
Power Series
1 1+
2
+ 1
3
+ 1
4
+..
For x = | r | > 1, the general terms of P(r)
P(r) = 1 + r r2
+
2
+
3
+
4
+.. diverges.r3 r4
P(–1) = 1 – is basically the
alternating harmonic series. It converges conditionally.
1 1 –
2
+ 1
3
+ 1
4 – ..
For x = ±1:
rn
n ∞ as n  ∞ (verify this by L'Hopital's rule).
Hence
We summarize the result using the following picture:
0
–1 converges (absolutely) 1 divergesdiverges
P(x) = Σ
xn
n

22. P(1) = 1 + is basically the
harmonic series, so it diverges.
Power Series
1 1+
2
+ 1
3
+ 1
4
+..
For x = | r | > 1, the general terms of P(r)
P(r) = 1 + r r2
+
2
+
3
+
4
+.. diverges.r3 r4
P(–1) = 1 – is basically the
alternating harmonic series. It converges conditionally.
1 1 –
2
+ 1
3
+ 1
4 – ..
For x = ±1:
rn
n ∞ as n  ∞ (verify this by L'Hopital's rule).
Hence
We summarize the result using the following picture:
0
–1 converges (absolutely) 1
converges conditionally diverges
divergesdiverges
P(x) = Σ
xn
n

23. Theorem: Given a power series , one of the
following must be true:
Power Series
Σn=0
P(x)
∞

24. Theorem: Given a power series , one of the
following must be true:
I. P(x) converges at x = 0, diverges everywhere else.
Power Series
ΣP(x)
∞
n=0

25. Theorem: Given a power series , one of the
following must be true:
I. P(x) converges at x = 0, diverges everywhere else.
Power Series
ΣP(x)
∞
0
converges at 0 divergesdiverges
n=0
x

26. Theorem: Given a power series , one of the
following must be true:
I. P(x) converges at x = 0, diverges everywhere else.
Power Series
ΣP(x)
∞
II. P(x) converges (absolutely) for | x | < R for some
positive R,
0
converges at 0 divergesdiverges
n=0
x

27. Theorem: Given a power series , one of the
following must be true:
I. P(x) converges at x = 0, diverges everywhere else.
Power Series
ΣP(x)
∞
II. P(x) converges (absolutely) for | x | < R for some
positive R, diverges for | x | > R,
0
converges at 0 divergesdiverges
n=0
converges (absolutely) R divergesdiverges
x
x
0
–R

28. Theorem: Given a power series , one of the
following must be true:
I. P(x) converges at x = 0, diverges everywhere else.
Power Series
ΣP(x)
∞
II. P(x) converges (absolutely) for | x | < R for some
positive R, diverges for | x | > R,
and at x = ±R, the series may converge or diverge.
0
converges at 0 divergesdiverges
n=0
converges (absolutely) R divergesdiverges
x
x
0? ?
–R

29. Theorem: Given a power series , one of the
following must be true:
I. P(x) converges at x = 0, diverges everywhere else.
Power Series
ΣP(x)
∞
II. P(x) converges (absolutely) for | x | < R for some
positive R, diverges for | x | > R,
and at x = ±R, the series may converge or diverge.
R is called the radius of convergence.
0
converges at 0 divergesdiverges
0
–R converges (absolutely) R
? ?
divergesdiverges
n=0
x
x

30. Theorem: Given a power series , one of the
following must be true:
I. P(x) converges at x = 0, diverges everywhere else.
Power Series
ΣP(x)
∞
II. P(x) converges (absolutely) for | x | < R for some
positive R, diverges for | x | > R,
and at x = ±R, the series may converge or diverge.
R is called the radius of convergence.
0
converges at 0 divergesdiverges
0
converges (absolutely) R
? ?
divergesdiverges
III. P(x) converges everywhere.
0
convergesconverges
n=0
x
x
x–R

31. Power Series
We may calculate the radius of convergence using
the ratio test or the root test.

32. Power Series
We may calculate the radius of convergence using
the ratio test or the root test.
Example C.
Discuss the convergence of P(x) = Σn=0
∞
xn
n! .

33. Power Series
Since the coefficients are factorials, use the ratio test:
We may calculate the radius of convergence using
the ratio test or the root test.
Example C.
Discuss the convergence of P(x) = Σn=0
∞
xn
n! .

34. Power Series
Since the coefficients are factorials, use the ratio test:
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 ... convergesP(x) =
for x if |an+1xn+1|*
|anxn|
1
< 1 as n  ∞.
We may calculate the radius of convergence using
the ratio test or the root test.
Example C.
Discuss the convergence of P(x) = Σn=0
∞
xn
n! .

35. Power Series
Since the coefficients are factorials, use the ratio test:
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 ... convergesP(x) =
for x if |an+1xn+1|*
|anxn|
1
< 1 as n  ∞.
|an+1xn+1|*
|anxn|
1
=
xn+1
(n+1)!
*
n!
xn
We may calculate the radius of convergence using
the ratio test or the root test.
Example C.
Discuss the convergence of P(x) = Σn=0
∞
xn
n! .

36. Power Series
Since the coefficients are factorials, use the ratio test:
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 ... convergesP(x) =
for x if |an+1xn+1|*
|anxn|
1
< 1 as n  ∞.
|an+1xn+1|*
|anxn|
1
=
xn+1
(n+1)!
*
n!
xn =
x
n+1
We may calculate the radius of convergence using
the ratio test or the root test.
Example C.
Discuss the convergence of P(x) = Σn=0
∞
xn
n! .

37. Power Series
Since the coefficients are factorials, use the ratio test:
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 ... convergesP(x) =
for x if |an+1xn+1|*
|anxn|
1
< 1 as n  ∞.
|an+1xn+1|*
|anxn|
1
=
xn+1
(n+1)!
*
n!
xn =
x
n+1
We may calculate the radius of convergence using
the ratio test or the root test.
As n  ∞, for any real number x,
x
n+1  0 < 1.
Example C.
Discuss the convergence of P(x) = Σn=0
∞
xn
n! .

38. Power Series
Since the coefficients are factorials, use the ratio test:
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 ... convergesP(x) =
for x if |an+1xn+1|*
|anxn|
1
< 1 as n  ∞.
|an+1xn+1|*
|anxn|
1
=
xn+1
(n+1)!
*
n!
xn =
x
n+1
We may calculate the radius of convergence using
the ratio test or the root test.
As n  ∞, for any real number x,
x
n+1  0 < 1.
Example C.
Discuss the convergence of P(x) = Σn=0
∞
xn
n! .

39. Power Series
Since the coefficients are factorials, use the ratio test:
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 ... convergesP(x) =
for x if |an+1xn+1|*
|anxn|
1
< 1 as n  ∞.
|an+1xn+1|*
|anxn|
1
=
xn+1
(n+1)!
*
n!
xn =
x
n+1
Example C.
Discuss the convergence of P(x) = Σn=0
∞
xn
n! .
We may calculate the radius of convergence using
the ratio test or the root test.
As n  ∞, for any real number x,
x
n+1  0 < 1.
Hence the series converegs for all numbers x.

40. Example D.
Discuss the convergence of P(x) =
Power Series
Σn=0
∞
xn
2n

41. Power Series
By the root test, for any real number x,
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 + a4x4 ….P(x) =
converges if lim |anxn| < 1, diverges if it's > 1.n
n∞
Example D.
Discuss the convergence of P(x) = Σn=0
∞
xn
2n

42. Power Series
By the root test, for any real number x,
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 + a4x4 ….P(x) =
converges if lim |anxn| < 1, diverges if it's > 1.
anxn = xn
2n,
n
n∞
Example D.
Discuss the convergence of P(x) = Σn=0
∞
xn
2n

43. Power Series
By the root test, for any real number x,
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 + a4x4 ….P(x) =
converges if lim |anxn| < 1, diverges if it's > 1.
anxn = xn
2n,
so the series converges if
n
n
n∞
lim |xn/2n|n∞
Example D.
Discuss the convergence of P(x) = Σn=0
∞
xn
2n

44. Power Series
By the root test, for any real number x,
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 + a4x4 ….P(x) =
converges if lim |anxn| < 1, diverges if it's > 1.
anxn = xn
2n,
so the series converges if
n
n
n∞
lim |xn/2n| = |x/2| < 1n∞
Example D.
Discuss the convergence of P(x) = Σn=0
∞
xn
2n

45. Power Series
By the root test, for any real number x,
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 + a4x4 ….P(x) =
converges if lim |anxn| < 1, diverges if it's > 1.
anxn = xn
2n,
so the series converges if
n
n
n∞
lim |xn/2n| = |x/2| < 1 or | x | < 2.n∞
Example D.
Discuss the convergence of P(x) = Σn=0
∞
xn
2n

46. Power Series
By the root test, for any real number x,
Σn=0
∞
anxn = a0 + a1x + a2x2 + a3x3 + a4x4 ….P(x) =
converges if lim |anxn| < 1, diverges if it's > 1.
anxn = xn
2n,
and diverges if | x | > 2,
and the radius of convergence R = 2.
so the series converges if
n
n
n∞
lim |xn/2n| = |x/2| < 1 or | x | < 2.n∞
Example D.
Discuss the convergence of P(x) = Σn=0
∞
xn
2n

47. Power Series
Finally, we clarify what happen at the boundary points
x = ±2.

48. Power Series
Finally, we clarify what happen at the boundary points
x = ±2.
P(2) = Σn=0
2n
2n = 1 + 1 + 1 + … so it diverges at x = 2.

49. Power Series
Finally, we clarify what happen at the boundary points
x = ±2.
P(2) = Σn=0
2n
2n = 1 + 1 + 1 + … so it diverges at x = 2.
P(–2) =Σn=0
(–2)n
2n
= 1 – 1 + 1 – 1.. so it diverges at x = –2.

50. Power Series
Finally, we clarify what happen at the boundary points
x = ±2.
P(2) = Σn=0
2n
2n = 1 + 1 + 1 + … so it diverges at x = 2.
P(–2) =Σn=0
(–2)n
2n
= 1 – 1 + 1 – 1.. so it diverges at x = –2.
So P(x) converges for | x | < 2, diverges for | x | ≥ 2.

51. Power Series
Finally, we clarify what happen at the boundary points
x = ±2.
P(2) = Σn=0
2n
2n = 1 + 1 + 1 + … so it diverges at x = 2.
P(–2) =Σn=0
(–2)n
2n
= 1 – 1 + 1 – 1.. so it diverges at x = –2.
So P(x) converges for | x | < 2, diverges for | x | ≥ 2.
We may shift a power series by replacing x by (x – a).

52. Power Series
Finally, we clarify what happen at the boundary points
x = ±2.
P(2) = Σn=0
2n
2n = 1 + 1 + 1 + … so it diverges at x = 2.
P(–2) =Σn=0
(–2)n
2n
= 1 – 1 + 1 – 1.. so it diverges at x = –2.
So P(x) converges for | x | < 2, diverges for | x | ≥ 2.
We may shift a power series by replacing x by (x – a).
Σn=0
∞
an(x – a)n = a0 + a1(x – a) + a2(x – a)2 + a3(x – a)3 +..
∞
is said to be a power series centered at x = a.
A power defined in the form of

53. Power Series
Finally, we clarify what happen at the boundary points
x = ±2.
P(2) = Σn=0
2n
2n = 1 + 1 + 1 + … so it diverges at x = 2.
P(–2) =Σn=0
(–2)n
2n
= 1 – 1 + 1 – 1.. so it diverges at x = –2.
So P(x) converges for | x | < 2, diverges for | x | ≥ 2.
We may shift a power series by replacing x by (x – a).
Σn=0
∞
an(x – a)n = a0 + a1(x – a) + a2(x – a)2 + a3(x – a)3 +..
The power series Σn=0
∞
anxn
is said to be a power series centered at x = a.
is the special case of
Σn=0
∞
an(x – a)n that’s center at a = 0.
A power defined in the form of

54. Theorem: Given a power series , one of the
following must be true:
Power Series
Σn=0
∞
an(x – a)n

55. Theorem: Given a power series , one of the
following must be true:
I. It converges at x = a, diverges everywhere else.
Power Series
a
converges at a divergesdiverges
Σn=0
∞
an(x – a)n

56. Theorem: Given a power series , one of the
following must be true:
I. It converges at x = a, diverges everywhere else.
Power Series
II. For some positive R, it converges (absolutely) for
| x – a | < R, diverges for | x – a | > R,
and at x = a±R, the series may converge or diverge.
(a – R, a + R) is called the interval of convergence.
a
converges at a divergesdiverges
a
a – R converges (absolutely) a + R
? ?
divergesdiverges
Σn=0
∞
an(x – a)n

57. Theorem: Given a power series , one of the
following must be true:
I. It converges at x = a, diverges everywhere else.
Power Series
II. For some positive R, it converges (absolutely) for
| x – a | < R, diverges for | x – a | > R,
and at x = a±R, the series may converge or diverge.
(a – R, a + R) is called the interval of convergence.
a
converges at a divergesdiverges
a
a – R converges (absolutely) a + R
? ?
divergesdiverges
III. It converges everywhere.
a
convergesconverges
Σn=0
∞
an(x – a)n

58. Power Series
Example E.
Discuss the convergence of P(x) = Σn=1
∞
Ln(n)(x – 3)n
n

59. Power Series
Use the ratio test,
Example E.
Discuss the convergence of P(x) = Σn=1
∞
Ln(n)(x – 3)n
n

60. Power Series
Use the ratio test,
|an+1 (x –a)n+1|*
|an(x – a)n|
1
Example E.
Discuss the convergence of P(x) = Σn=1
∞
Ln(n)(x – 3)n
n

61. Example E.
Discuss the convergence of P(x) =
Power Series
Σn=1
∞
Ln(n)(x – 3)n
n
Use the ratio test,
|an+1 (x –a)n+1|*
|an(x – a)n|
1
= Ln(n+1)(x – 3)n+1
n+1 *
Ln(n)(x – 3)n
n

62. Power Series
Use the ratio test,
|an+1 (x –a)n+1|*
|an(x – a)n|
1
= Ln(n+1)(x – 3)n+1
n+1 *
Ln(n)(x – 3)n
n
= n * Ln(n+1) (x – 3)
(n+1) * Ln(n)
Example E.
Discuss the convergence of P(x) = Σn=1
∞
Ln(n)(x – 3)n
n

63. Power Series
Use the ratio test,
|an+1 (x –a)n+1|*
|an(x – a)n|
1
= Ln(n+1)(x – 3)n+1
n+1 *
Ln(n)(x – 3)n
n
= n * Ln(n+1) (x – 3)
(n+1) * Ln(n)
, as n  ∞
Example E.
Discuss the convergence of P(x) = Σn=1
∞
Ln(n)(x – 3)n
n

64. Power Series
Use the ratio test,
|an+1 (x –a)n+1|*
|an(x – a)n|
1
= Ln(n+1)(x – 3)n+1
n+1 *
Ln(n)(x – 3)n
n
= n * Ln(n+1) (x – 3)
(n+1) * Ln(n)
, as n  ∞
1 1
Example E.
Discuss the convergence of P(x) = Σn=1
∞
Ln(n)(x – 3)n
n

65. Power Series
Use the ratio test,
|an+1 (x –a)n+1|*
|an(x – a)n|
1
= Ln(n+1)(x – 3)n+1
n+1 *
Ln(n)(x – 3)n
n
= n * Ln(n+1) (x – 3)
(n+1) * Ln(n)
, as n  ∞ we get |x – 3|
1 1
Example E.
Discuss the convergence of P(x) = Σn=1
∞
Ln(n)(x – 3)n
n

66. Power Series
Use the ratio test,
|an+1 (x –a)n+1|*
|an(x – a)n|
1
= Ln(n+1)(x – 3)n+1
n+1 *
Ln(n)(x – 3)n
n
= n * Ln(n+1) (x – 3)
(n+1) * Ln(n)
, as n  ∞ we get |x – 3|
1 1
The series converges if |x – 3| < 1
Example E.
Discuss the convergence of P(x) = Σn=1
∞
Ln(n)(x – 3)n
n

67. Power Series
Use the ratio test,
|an+1 (x –a)n+1|*
|an(x – a)n|
1
= Ln(n+1)(x – 3)n+1
n+1 *
Ln(n)(x – 3)n
n
= n * Ln(n+1) (x – 3)
(n+1) * Ln(n)
, as n  ∞ we get |x – 3|
1 1
The series converges if |x – 3| < 1 so the interval
of convergence is (2, 4).
Example E.
Discuss the convergence of P(x) = Σn=1
∞
Ln(n)(x – 3)n
n

68. Power Series
For the end points of the interval:

69. P(4) =
Power Series
Σn=1
∞ Ln(n)(4 – 3)n
n
For the end points of the interval:

70. P(4) =
Power Series
Σn=1
∞ Ln(n)(4 – 3)n
n = Σn=1
∞
Ln(n)
n
For the end points of the interval:

71. P(4) =
Power Series
Σn=1
∞ Ln(n)(4 – 3)n
n = Σn=1
∞
Ln(n)
n > Σn=2
∞
1
n = ∞
For the end points of the interval:

72. P(4) =
Power Series
Σn=1
∞ Ln(n)(4 – 3)n
n
Σn=1
∞ Ln(n)(2 – 3)n
n
= Σn=1
∞
Ln(n)
n > Σn=2
∞
1
n = ∞
P(2) =
For the end points of the interval:

73. P(4) =
Power Series
Σn=1
∞ Ln(n)(4 – 3)n
n
Σn=1
∞ Ln(n)(2 – 3)n
n
= Σn=1
∞
Ln(n)
n > Σn=2
∞
1
n = ∞
P(2) =
For the end points of the interval:
= Σn=1
∞
(–1)nLn(n)
n

74. P(4) =
Power Series
Σn=1
∞ Ln(n)(4 – 3)n
n
Σn=1
∞ Ln(n)(2 – 3)n
n
= Σn=1
∞
Ln(n)
n > Σn=2
∞
1
n = ∞
P(2) =
For the end points of the interval:
= Σn=1
∞
(–1)nLn(n)
n
is a decreasing alternating series with term  0,
so it converges.

75. P(4) =
Power Series
Σn=1
∞ Ln(n)(4 – 3)n
n
Σn=1
∞ Ln(n)(2 – 3)n
n
= Σn=1
∞
Ln(n)
n
is a decreasing alternating series with term  0,
so it converges. However it converges conditionally.
> Σn=2
∞
1
n = ∞
P(2) =
For the end points of the interval:
= Σn=1
∞
(–1)nLn(n)
n

76. P(4) =
Power Series
Σn=1
∞ Ln(n)(4 – 3)n
n
Σn=1
∞ Ln(n)(2 – 3)n
n
= Σn=1
∞
Ln(n)
n > Σn=2
∞
1
n = ∞
P(2) =
For the end points of the interval:
= Σn=1
∞
(–1)nLn(n)
n
Hence we have the following graph for convergence:
3
2 converges (absolutely) 4
converges conditionally
divergesdiverges
diverges
is a decreasing alternating series with term  0,
so it converges. However it converges conditionally.