internal assessment

Kohshi Gokita

March 28, 2010

IB Math SL. Per. 8

Internal AssessmentInfinite Summation

The infinite sequence below is considered

t 0=1 t 1=(xlna )01

1 ! t 2=

(xlna )2

2! …………..

So, the general term of this sequence is expressed as follows;

t n=( xlna)n

n!

The sum T n(a,x) of the first (n+1) terms of the above sequence is considered

T n(a ,x) = t 0 + t 1 + t 2 + t 3 + …………….+t n

= ∑k=0

n

tk

=∑k=0

n (xlna)k

k !

Aim;

In this task, the sum of infinite sequence t n is considered, hence, the value of T n

(a, x) as a n approaches∞,

limn → ∞

T n(a , x)

would be calculated

I. In this part of investigation, the value of x is fixed on 1 and the effect of change in the value of a over the value of T n(a, x) would be observed by numerical simulation.

1. Consider the following sequence of terms where x=1 and a=2. And calculate the

sum T n of the first n terms of the above sequence for 0≤n≤10.

1, + ( ln 2 )❑

1 +

( ln 2 )2

2∗1 +

( ln2 )3

3∗2∗1 + ………..

0≤n≤10:

a x n t T

2 1 0 1 1

1 0.693147 1.693147

2 0.240227 1.933374

3 0.055504 1.988878

4 0.009618 1.998496

5 0.001333 1.999829

6 0.000154 1.999983

7 1.53E-05 1.999999

8 1.32E-06 2

9 1.02E-07 2

10 7.05E-09 2

0 2 4 6 8 10 120

0.5

1

1.5

2

2.5

0≤n≤20:

a x n t T

2 1 0 1 1

1 0.693147 1.693147

2 0.240227 1.933374

3 0.055504 1.988878

4 0.009618 1.998496

5 0.001333 1.999829

6 0.000154 1.999983

7 1.53E-05 1.999999

8 1.32E-06 2

9 1.02E-07 2

10 7.05E-09 2

11 4.45E-10 2

12 2.57E-11 2

13 1.37E-12 2

14 6.78E-14 2

15 3.13E-15 2

16 1.36E-16 2

17 5.53E-18 2

18 2.13E-19 2

19 7.77E-21 2

20 2.69E-22 2

0 5 10 15 20 250

0.5

1

1.5

2

2.5

*According to data, as n increases, sum of infinity approaches to the value of a. In this

case, the sum gets closed to 2.



1, + ( ln 3 )❑

1 +

(ln 3 )2

2∗1 +

( ln3 )3

3∗2∗1 + ………..

0≤n≤10:

a x n t T

3 1 0 1 1

1 1.098612 2.098612

2 0.603474 2.702087

3 0.220995 2.923082

4 0.060697 2.983779

5 0.013336 2.997115

6 0.002442 2.999557

7 0.000383 2.99994

8 5.26E-05 2.999993

9 6.42E-06 2.999999

10 7.06E-07 3

0 2 4 6 8 10 120

0.5

1

1.5

2

2.5

3

3.5

0≤n≤20:

a x n t T

3 1 0 1 1

1 1.098612 2.098612

2 0.603474 2.702087

3 0.220995 2.923082

4 0.060697 2.983779

5 0.013336 2.997115

6 0.002442 2.999557

7 0.000383 2.99994

8 5.26E-05 2.999993

9 6.42E-06 2.999999

10 7.06E-07 3

11 7.05E-08 3

12 6.45E-09 3

13 5.45E-10 3

14 4.28E-11 3

15 3.13E-12 3

16 2.15E-13 3

17 1.39E-14 3

18 8.49E-16 3

19 4.91E-17 3

20 2.7E-18 3

0 5 10 15 20 250

0.5

1

1.5

2

2.5

3

3.5





1, + (ln 4 )❑

1 +

( ln 4 )2

2∗1 +

(ln 4 )3

3∗2∗1 + ………..

0≤n≤10:

a x n t T

4 1 0 1 1

1 1.386294 2.386294

2 0.960906 3.3472

3 0.444033 3.791233

4 0.15389 3.945123

5 0.042667 3.987791

6 0.009858 3.997649

7 0.001952 3.999601

8 0.000338 3.99994

9 5.21E-05 3.999992

10 7.22E-06 3.999999

0 2 4 6 8 10 120

0.5

1

1.5

2

2.5

3

3.5

4

4.5

Series2

0≤n≤20:

a x n t T

4 1 0 1 1

1 1.386294 2.386294

2 0.960906 3.3472

3 0.444033 3.791233

4 0.15389 3.945123

5 0.042667 3.987791

6 0.009858 3.997649

7 0.001952 3.999601

8 0.000338 3.99994

9 5.21E-05 3.999992

10 7.22E-06 3.999999

11 9.1E-07 4

12 1.05E-07 4

13 1.12E-08 4

14 1.11E-09 4

15 1.03E-10 4

16 8.89E-12 4

17 7.25E-13 4

18 5.59E-14 4

19 4.08E-15 4

20 2.82E-16 4

0 5 10 15 20 250

0.5

1

1.5

2

2.5

3

3.5

4

4.5



4. Now, consider a general sequence where x=1. And calculate the sum T n of the

first n terms of the above sequence for 0≤n≤10 for different values of a.

1, + ( lna)❑

1 +

( lna)2

2∗1 +

(lna )3

3∗2∗1 + ………..

If x=1 and a=10, the equation will be

1, + (ln 10 )❑

1 +

( ln 10 )2

2∗1 +

( ln10 )3

3∗2∗1 + ………..

0≤n≤10:

a x n t T

10 1 0 1 1

1 2.302585 3.302585

2 2.650949 5.953534

3 2.034679 7.988213

4 1.171255 9.159468

5 0.539383 9.698851

6 0.206996 9.905847

7 0.068089 9.973936

8 0.019598 9.993534

9 0.005014 9.998548

10 0.001154 9.999702

0 2 4 6 8 10 120

2

4

6

8

10

12

0≤n≤20:

a x n t T

10 1 0 1 1

1 2.302585 3.302585

2 2.650949 5.953534

3 2.034679 7.988213

4 1.171255 9.159468

5 0.539383 9.698851

6 0.206996 9.905847

7 0.068089 9.973936

8 0.019598 9.993534

9 0.005014 9.998548

10 0.001154 9.999702

11 0.000242 9.999944

12 4.64E-05 9.99999

13 8.21E-06 9.999998

14 1.35E-06 10

15 2.07E-07 10

16 2.98E-08 10

17 4.04E-09 10

18 5.17E-10 10

19 6.27E-11 10

20 7.21E-12 10

0 5 10 15 20 250

2

4

6

8

10

12



According to whole observation of experiment, it is proved that as n increases, the

sum approaches to value of a. In addition, as a increases, the converge speed gets slow

and slow. In cases of a=2 and x=1, sum becomes 2 when n is 8. However, in case of

a=10 and x=1, sum becomes 10 when n is 14. Ultimately, the general statement of the

infinite summation is

∑n=0

∞ ( lna)n

n!=a

II. In this part of investigation, how the change in the positive value of x as well as the positive value of a would affect the value of T n(a, x) would be observed by numerical simulation.



1, + (5 ln 2 )❑

1 +

(5 ln 2 )2

2∗1 +

(5 ln 2 )3

3∗2∗1 + ………..

0≤n≤10:

a x n t T

2 5 0 1 1

1 3.465736 4.465736

2 6.005663 10.4714

3 6.938014 17.40941

4 6.011331 23.42074

5 4.166737 27.58748

6 2.406802 29.99428

7 1.19162 31.1859

8 0.51623 31.70213

9 0.198791 31.90092

10 0.068896 31.96982

0 2 4 6 8 10 120

5

10

15

20

25

30

35

0≤n≤20;

a x n t T

2 5 0 1 1

1 3.465736 4.465736

2 6.005663 10.4714

3 6.938014 17.40941

4 6.011331 23.42074

5 4.166737 27.58748

6 2.406802 29.99428

7 1.19162 31.1859

8 0.51623 31.70213

9 0.198791 31.90092

10 0.068896 31.96982

11 0.021707 31.99152

12 0.006269 31.99779

13 0.001671 31.99946

14 0.000414 31.99988

15 9.56E-05 31.99997

16 2.07E-05 31.99999

17 4.22E-06 32

18 8.13E-07 32

19 1.48E-07 32

20 2.57E-08 32

0 5 10 15 20 250

5

10

15

20

25

30

35

*According to data, the graph’s shape changes. In this case, sum becomes 32 when n is

16. In addition, in the graph, there is a inflection point. The curve of the graph changes.

a=2, x=5 limn → ∞

T n(2 ,5) = 25 = 32



1, + (3 ln 2 )❑

1 +

(3 ln 2 )2

2∗1 +

(3 ln 2 )3

3∗2∗1 + ………..

0≤n≤10:

a x n t T

2 3 0 1 1

12.0794

42

3.0794

42

22.1620

39

5.2414

8

31.4986

11

6.7400

91

40.7790

68

7.5191

59

50.3240

05

7.8431

65

60.1122

92

7.9554

57

70.0333

58

7.9888

14

80.0086

71

7.9974

85

90.0020

03

7.9994

88

100.0004

17

7.9999

05

0 2 4 6 8 10 120

1

2

3

4

5

6

7

8

9

0≤n≤20:

a x n t T

2 3 0 1 1

12.0794

42

3.0794

42

22.1620

39

5.2414

8

31.4986

11

6.7400

91

40.7790

68

7.5191

59

50.3240

05

7.8431

65

60.1122

92

7.9554

57

70.0333

58

7.9888

14

80.0086

71

7.9974

85

90.0020

03

7.9994

88

100.0004

17

7.9999

05

117.88E-

05

7.9999

84

121.36E-

05

7.9999

97

132.18E-

068

143.24E-

078

154.49E-

088

165.84E-

098

17 7.15E- 8

10

188.25E-

118

199.03E-

128

209.39E-

138

0 5 10 15 20 250

1

2

3

4

5

6

7

8

9

*According to data, the graph’s shape changes. In this case, sum becomes 8 when n is

about5. In addition, in the graph, there is a inflection point. The curve of the graph

changes.

a=2, x=3 limn → ∞

T n(2 ,3) = 23 = 8

According to whole observation of investigation, it is proved that if the value of x

changes, the sum also changes. However, it is not that the sum approaches value of a.

The sum approaches the number of ax. In investigation in part I, the sum becomes the

same value as a but it is that because x is fixed on 1. In the case of a=2 and x=5, the

total sum becomes 25, 32. Ultimately, the general statement of the infinite summation is

∑n=0

∞ ( lna)n

n!=ax

III. In this part of investigation, how the negative value of x as well as the positive value of a would affect the value of T n(a, x) would be observed by numerical simulation. In addition, how the change in the positive value of a as well as the value of x would affect the value of T n(a, x) would be observed by numerical simulation.

1. Consider the following sequence of terms where x= -2 and a=2. And calculate the


1, + (−2 ln 2 )❑

1 +

(−2 ln 2 )2

2∗1 +

(−2 ln 2 )3

3∗2∗1 + ………..

0≤ n ≤10

a x n t T

2 -2 0 1 1

1-1.38629-0.38629

20.9609060.574612

3-0.444030.130579

4 0.153890.284469

5-0.042670.241801

60.009858 0.25166

7-0.001950.249707

80.0003380.250046

9-5.2E-050.249994

107.22E-060.250001

0 2 4 6 8 10 12

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

Series2

0≤ n ≤20

a x n t T

2 -2 0 1 1

1 -1.38629 -0.38629

2 0.960906 0.574612

3 -0.44403 0.130579

4 0.15389 0.284469

5 -0.04267 0.241801

6 0.009858 0.25166

7 -0.00195 0.249707

8 0.000338 0.250046

9 -5.2E-05 0.249994

10 7.22E-06 0.250001

11 -9.1E-07 0.25

12 1.05E-07 0.25

13 -1.1E-08 0.25

14 1.11E-09 0.25

15 -1E-10 0.25

16 8.89E-12 0.25

17 -7.3E-13 0.25

18 5.59E-14 0.25

19 -4.1E-15 0.25

20 2.82E-16 0.25

0 5 10 15 20 25

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

Series2

*According to the graph, the shape has dramatically changed from the graph of 0≤ x.

When n is 2, the sum value becomes negative but when n is 3, the value goes to positive

again. However, when n is 4, the sum value does not go to negative but it decreases

once again. From n becomes 4, the sum value constantly becomes 0.25.

a=2, x=-2 limn → ∞

T n(2 ,−2) = 2−2 = 0.25

2. Consider the following sequence of terms where x= -2 and a= 1/2. And calculate the


1, + (−2 ln ( 12))

❑

1 + (−2 ln ( 1

2))

2

2∗1

+ (−2 ln ( 12))

3

3∗2∗1

+ ………..

0≤n≤10

a x n t T

0.5 -2 0 1 1

11.3862942.386294

20.960906 3.3472

30.4440333.791233

4 0.153893.945123

50.0426673.987791

60.0098583.997649

70.0019523.999601

80.000338 3.99994

95.21E-053.999992

107.22E-063.999999

0 2 4 6 8 10 120

0.5

1

1.5

2

2.5

3

3.5

4

4.5

Series2

0≤n≤20a x n t T

2 -2 0 1 1

1 -1.38629 -0.38629

2 0.960906 0.574612

3 -0.44403 0.130579

4 0.15389 0.284469

5 -0.04267 0.241801

6 0.009858 0.25166

7 -0.00195 0.249707

8 0.000338 0.250046

9 -5.2E-05 0.249994

10 7.22E-06 0.250001

11 -9.1E-07 0.25

12 1.05E-07 0.25

13 -1.1E-08 0.25

14 1.11E-09 0.25

15 -1E-10 0.25

16 8.89E-12 0.25

17 -7.3E-13 0.25

18 5.59E-14 0.25

19 -4.1E-15 0.25

20 2.82E-16 0.25

0 5 10 15 20 250

0.5

1

1.5

2

2.5

3

3.5

4

4.5

Series2

*According to the graph, the shape becomes back to the first type of graph. Although

the x is negative number, the graph shapes same looking as the graph of T n(a , 1). It might be because of a is half of 1.

a=1/2, x= -2 limn → ∞

T n(12

,−2) = ( 12)−2

= 4

3. Consider the following sequence of terms where x=2 and a=1/2. And calculate the


1, + (2 ln (12))

❑

1 + (2 ln (1

2))

2

2∗1

+ (2 ln (12))

3

3∗2∗1

+ ………..

0≤n≤10

a x n t T

0.5 2 0 1 1

1-

1.38629

-

0.38629

20.96090

6

0.57461

2

3-

0.44403

0.13057

9

4 0.153890.28446

9

5-

0.04267

0.24180

1

60.00985

80.25166

7-

0.00195

0.24970

7

80.00033

8

0.25004

6

9-5.2E-

05

0.24999

4

107.22E-

06

0.25000

1

0 2 4 6 8 10 12

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

Series2

0≤n≤20a x n t T

0.5 2 0 1 1

1-1.38629-0.38629

20.9609060.574612

3-0.444030.130579

4 0.153890.284469

5-0.042670.241801

60.009858 0.25166

7-0.001950.249707

80.0003380.250046

9-5.2E-050.249994

107.22E-060.250001

11-9.1E-07 0.25

121.05E-07 0.25

13-1.1E-08 0.25

141.11E-09 0.25

15 -1E-10 0.25

168.89E-12 0.25

17-7.3E-13 0.25

185.59E-14 0.25

19-4.1E-15 0.25

202.82E-16 0.25

0 5 10 15 20 25

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

Series2

*According to graph, it becomes exactly same as the graph of T n(2 ,−2) although this

graph’s x value is not negative. This graph is T n(12

,2), but it becomes exact same

looking graph. When n is 2, the sum value becomes negative but when n is 3, the value

goes to positive again. However, when n is 4, the sum value does not go to negative but

it decreases once again. From n becomes 4, the sum value constantly becomes 0.25.

a=1/2, x=2 limn → ∞

T n(12

,2) = 12

2

= 0.25

According to whole observation of investigation, it is proved that the sum value

follows the general statement of ∑n=0

∞ ( lna)n

n!=ax although value of x and a become

rational.

IV. In this part of investigation, how the rational value of x as well as the rational value of a such as 1/3, would affect the value of T n(a, x) would be observed by numerical simulation. In addition, how the irrational value of x as well as the irrational value of x such as π or √ 3 would affect the value of T n(a, x) would be observed by numerical simulation.

1. Consider the following sequence of terms where x=1/3, and a=1/3. And calculate the

sum T n of the first n terms of the above sequence for 0≤n≤10 and 0≤n≤20.

1, + ( 13

ln ( 13))

❑

1 + ( 1

3ln ( 1

3))

2

2∗1

+ ( 13

ln ( 123

))3

3∗2∗1

+ ………..

0≤n≤10

a x n t T

0.3333

33

0.3333

330 1 1

1 -0.36620.6337

96

20.0670

53

0.7008

49

3

-

0.0081

8

0.6926

64

40.0007

49

0.6934

13

5-5.5E-

05

0.6933

58

63.35E-

06

0.6933

61

7-1.8E-

07

0.6933

61

88.02E-

09

0.6933

61

9-3.3E-

10

0.6933

61

10 1.2E-110.6933

61

0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

1.2

Series2

0≤n≤20

a x n t T

0.3333

33

0.3333

330 1 1

1 -0.36620.6337

96

20.0670

53

0.7008

49

3

-

0.0081

8

0.6926

64

40.0007

49

0.6934

13

5-5.5E-

05

0.6933

58

63.35E-

06

0.6933

61

7-1.8E-

07

0.6933

61

88.02E-

09

0.6933

61

9-3.3E-

10

0.6933

61

10 1.2E-110.6933

61

11 -4E-130.6933

61

121.21E-

14

0.6933

61

13-3.4E-

16

0.6933

61

148.95E-

18

0.6933

61

15-2.2E-

19

0.6933

61

16 5E-210.6933

61

17-1.1E-

22

0.6933

61

182.19E-

24

0.6933

61

19-4.2E-

26

0.6933

61

20 7.73E- 0.6933

28 61

0 5 10 15 20 250

0.2

0.4

0.6

0.8

1

1.2

Series2

*According to graph, it also shapes like T n(2 ,−2) and T n(12

,2). But the difference is

that sum value never goes to negative although above both of sum value goes to

negative at least once. And from n=5, the sum value constantly get 0.693361.

a=1/3, x=1/3 limn → ∞

T n(13

,13) = 1

3

13 = 0.693361.

2. Consider the following sequence of terms where x=√ 3, and a=π. And calculate the


1, + ((√3 ) ln (π ))❑

1 + ((√3 ) ln (π ))2

2∗1 + ((√3 ) ln (π ))3

3∗2∗1 + ………..

0≤n≤10

a x n t T

π √3 0 1 1

11.9827

3

2.9827

3

21.9656

1

4.9483

4

31.2990

91

6.2474

31

40.6439

37

6.8913

68

50.2553

51

7.1467

19

60.0843

82

7.2311

01

70.0239

01

7.2550

02

80.0059

24

7.2609

25

90.0013

05

7.2622

3

100.0002

59

7.2624

89

0 2 4 6 8 10 120

1

2

3

4

5

6

7

8

Series2

0≤n≤20

a x n t T

3.1415

93

1.7320

510 1 1

11.9827

3

2.9827

3

21.9656

1

4.9483

4

31.2990

91

6.2474

31

40.6439

37

6.8913

68

50.2553

51

7.1467

19

60.0843

82

7.2311

01

70.0239

01

7.2550

02

80.0059

24

7.2609

25

90.0013

05

7.2622

3

100.0002

59

7.2624

89

114.66E-

05

7.2625

36

127.71E-

06

7.2625

43

131.18E-

06

7.2625

45

141.66E-

07

7.2625

45

15 2.2E-087.2625

45

162.73E-

09

7.2625

45

173.18E-

10

7.2625

45

18 3.5E-117.2625

45

193.66E-

12

7.2625

45

203.62E-

13

7.2625

45

0 5 10 15 20 250

1

2

3

4

5

6

7

8

Series2

*According to the graph, it becomes same shape as the graph of T n(2 ,1). When n

becomes 5, the sum value begins to constantly get 7.262545.

According to whole observation of investigation, it is proved that although value of x

and becomes something irrational, sum value can be defined by the general statement of

∑n=0

∞ ( lna)n

n!=a

Conclusion:According to all the observation of data, it can be defined that all of sum value follows

that general statement above. However, the shapes of graphing are not always same.

Some differs from others. From all the data that gathered, sum value of xlna can be

defined that

x a lna xlna

x¿0 a¿1 lna¿0 Positive +

0 2 4 6 8 10 120

10

20

30

40

x¿0 0¿a<¿1 lna<¿0 Negative

0 2 4 6 8 10 12-0.5

0

0.5

1

1.5

Series2

x¿0 a¿1 lna¿0 Negative

0 2 4 6 8 10 12-0.5

0

0.5

1

1.5

Series2

x¿0 0¿a<¿1 lna<¿0 Positive +

0 2 4 6 8 10 120

1

2

3

4

5

Series2

Ultimately, to answer to the aim, whatever the a and x are, infinite sum value follows

the general statement as n increases. So, t n is considered, hence, the value of T n (a, x) as

a n approaches∞,

limn → ∞

T n(a , x)

It can be conclude with

aa¿0

any positive real numbers.

xany real numbers.

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