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The purpose of this paper is to investigate an infinite summation patter where Ln(a) is a constant and the coefficient of x is an increasing factor to Ln(a).

Extracts from this document...

Introduction

Infinite Summation

Math IA

Consider the following sequence of terms where x=1 and a=2 under the terms that 0≤n≤10:

tn=

n	t(n)	S(n)
0.000000	1	1
1.000000	0.69314718	1.693147
2.000000	0.24022651	1.933373
3.000000	0.05550411	1.988877
4.000000	0.00961813	1.998495
5.000000	0.00133336	1.999829
6.000000	0.00015404	1.999983
7.000000	1.5253E-05	1.999998
8.000000	1.3215E-06	1.999999
9.000000	1.0178E-07	1.999999
10.000000	7.0549E-09	2

As n → +∞, Sn → +2

Consider the following sequence of terms where x=1 and a=3:

tn=

n	t(n)	S(n)
0.000000	1.000000	1.000000
1.000000	1.098612	2.098612
2.000000	0.603474	2.702087
3.000000	0.220995	2.923082
4.000000	0.060697	2.983779
5.000000	0.013336	2.997115
6.000000	0.002442	2.999557
7.000000	0.000383	2.999940
8.000000	0.000053	2.999993
9.000000	0.000006	2.999999
10.000000	0.000001	3.000000

As n → +∞, Sn → +3

There is a horizontal asymptote as n approaches positive infinite (∞). As n approaches positive infinite then Sn will approach positive three. Sn approaches a horizontal asymptote when y=3. There is a y-intercept at (0,1).

As n → +∞, Sn → +4

There is a horizontal asymptote as n approaches positive infinite (∞). As n approaches positive infinite then Sn will approach positive four. Sn approaches a horizontal asymptote when y=4.

Middle

5.666868

3.000000

0.277521

5.944389

4.000000

0.048091

5.992480

5.000000

0.006667

5.999146

6.000000

0.000770

5.999917

7.000000

0.000076

5.999993

8.000000

0.000007

5.999999

9.000000

0.000001

6.000000

As n → +∞, Sn → +6

Let a=2 and calculate various positive values for x:

x	t(n)	S(n)
0.0	1.000000	1.000000
1.0	0.693147	1.693147
2.0	0.480453	2.173600
3.0	0.166512	2.340113
4.0	0.038473	2.378585
5.0	0.006667	2.385252

In the graph, when x is approaching infinite the Sn values are increasing steadily. When various values are used for x then there is an exponential growth.

Let a=3 then calculate for various positive values of x:

x	t(n)	S(n)
0.0	1.000000	1.000000
1.0	1.098612	2.098612
2.0	1.206949	3.305561
3.0	0.662984	3.968546
4.0	0.242788	4.211333
5.0	0.066682	4.278016

In the graph, when x is approaching infinite the Sn values are increasing steadily. When various values are used for x then there is an exponential growth.

Evidence:

tn=

n	t(n)	S(n)
0.000000	1.000000	1.000000
1.000000	2.772589	3.772589
2.000000	0.960906	4.733495
3.000000	0.222016	4.955511
4.000000	0.038473	4.993984
5.000000	0.005333	4.999317
6.000000	0.000616	4.999933
7.000000	0.000061	4.999994
8.000000	0.000005	5.000000
9.000000	0.000000	5.000000

As n → +∞, Sn → +6

Sn approaches a horizontal asymptote when y=6. There is a y-intercept at (0,1).

tn =

n	t(n)	S(n)
0.000000	1.000000	1.000000
1.000000	4.158883	5.158883
2.000000	1.441359	6.600242
3.000000	0.333025	6.933267
4.000000	0.057709	6.990976
5.000000	0.008000	6.998976
6.000000	0.000924	6.999900
7.000000	0.000092	6.999991
8.000000	0.000008	6.999999
9.000000	0.000001	7.000000

As n → +∞, Sn → +7

There is a horizontal asymptote as n approaches positive infinite (∞). As n approaches positive infinite then Sn will approach positive seven. Sn approaches a horizontal asymptote when y=7. There is a y-intercept at (0,1).

tn =

n	t(n)	S(n)
0.000000	1.000000	1.000000
1.000000	4.852030	5.852030
2.000000	1.681586	7.533616
3.000000	0.388529	7.922145
4.000000	0.067327	7.989471
5.000000	0.009333	7.998805
6.000000	0.001078	7.999883
7.000000	0.000107	7.999990
8.000000	0.000009	7.999999
9.000000	0.000001	8.000000

As n → +∞, Sn → +8

There is a horizontal asymptote as n approaches positive infinite (∞). As n approaches positive infinite then Sn will approach positive eight. Sn approaches a horizontal asymptote when y=8. There is a y-intercept at (0,1).

tn =

n	t(n)	S(n)
0.000000	1.000000	1.000000
1.000000	5.545177	6.545177
2.000000	1.921812	8.466990
3.000000	0.444033	8.911022
4.000000	0.076945	8.987967
5.000000	0.010667	8.998634
6.000000	0.001232	8.999867
7.000000	0.000122	8.999989
8.000000	0.000011	8.999999
9.000000	0.000001	9.000000

As n → +∞, Sn → + 9

There is a horizontal asymptote as n approaches positive infinite (∞). As n approaches positive infinite then Sn will approach positive nine. Sn approaches a horizontal asymptote when y=9. There is a y-intercept at (0,1).

tn =

n	t(n)	S(n)
0.000000	1.000000	1.000000
1.000000	6.238325	7.238325
2.000000	2.162039	9.400363
3.000000	0.499537	9.899900
4.000000	0.086563	9.986463
5.000000	0.012000	9.998464
6.000000	0.001386	9.999850
7.000000	0.000137	9.999987
8.000000	0.000012	9.999999
9.000000	0.000001	10.000000
10.000000	0.000000	10.000000

Conclusion

tn =

n	t(n)	S(n)
0.000000	1.000000	1.000000
1.000000	5.493061	6.493061
2.000000	3.017372	9.510434
3.000000	1.104974	10.615408
4.000000	0.303485	10.918893
5.000000	0.066682	10.985575
6.000000	0.012210	10.997785
7.000000	0.001916	10.999701
8.000000	0.000263	10.999964
9.000000	0.000032	10.999996

As n → +∞, Sn → + 11

There is a horizontal asymptote as n approaches positive infinite (∞).

tn=

n	t(n)	S(n)
0.0	1.000000	1.000000
1.0	7.690286	8.690286
2.0	4.224321	12.914607
3.0	1.546964	14.461571
4.0	0.424878	14.886450
5.0	0.093355	14.979805
6.0	0.017094	14.996898
7.0	0.002683	14.999581
8.0	0.000368	14.999950
9.0	0.000045	14.999995

As n → +∞, Sn → + 15

There is a horizontal asymptote as n approaches positive infinite (∞). As n approaches positive infinite then Sn will approach positive fifteen. Sn approaches a horizontal asymptote when y=15. There is a y-intercept at (0,1).

tn=

n	t(n)	S(n)
0.0	1.000000	1.000000
1.0	8.788898	9.788898
2.0	4.827796	14.616694
3.0	1.767959	16.384653
4.0	0.485575	16.870228
5.0	0.106692	16.976920
6.0	0.019535	16.996455
7.0	0.003066	16.999521
8.0	0.000421	16.999942
9.0	0.000051	16.999994

As n → +∞, Sn → + 17

There is a horizontal asymptote as n approaches positive infinite (∞). Sn approaches a horizontal asymptote when y=17. There is a y-intercept at (0,1).

Checks with various numbers:

tn =

n	t(n)	S(n)
0.0	1.000000	1.000000
1.0	48.283137	49.283137
2.0	38.854356	88.137493
3.0	20.844558	108.982051
4.0	8.387005	117.369057
5.0	2.699673	120.068729
6.0	0.724159	120.792889
7.0	0.166498	120.959387
8.0	0.033496	120.992883
9.0	0.005990	120.998873
10.000000	0.000964	120.999837

tn =

n	t(n)	S(n)
0.0	1.000000	1.000000
1.0	13.961881	14.961881
2.0	32.489020	47.450901
3.0	50.400871	97.851772
4.0	58.640914	156.492686
5.0	54.582497	211.075184
6.0	42.337463	253.412647
7.0	28.148125	281.560772
8.0	16.375032	297.935804
9.0	8.467639	306.403443
10.000000	3.940806	310.344249

n	t(n)	S(n)
0.0	1.000000	1.000000
1.0	179.175947	180.175947
2.0	160.520100	340.696047
3.0	95.871136	436.567183
4.0	42.944504	479.511687
5.0	15.389244	494.900931
6.0	4.595637	499.496569
7.0	1.176325	500.672894
8.0	0.263461	500.936356
9.0	0.052451	500.988807
10.000000	0.009398	500.998205

tn =

Conclusion:

The limitation to this general statement is that the scope of the evidence is limited to only a few combinations of x and a .

The continuous observation of various graphs with different values used in place of a and x provide evidence for the general statement.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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