S = 1 + 1/(1+2) + 1/(1+2+3) + … + 1/(1+2+3+4+…+n)
Find a simple expression for S
What is the limit of S when n ==> infinity ?
clearvars, clc
den=1; s=1;
for new=2:50
den=den+new;
s=s+1/den;
fprintf('%3d %15.13f %15.13f %15.13f\n',new,1/den,s,2-s)
end
was used to tabulate the values:
n last term S 2 - S
2 0.3333333333333 1.3333333333333 0.6666666666667
3 0.1666666666667 1.5000000000000 0.5000000000000
4 0.1000000000000 1.6000000000000 0.4000000000000
5 0.0666666666667 1.6666666666667 0.3333333333333
6 0.0476190476190 1.7142857142857 0.2857142857143
7 0.0357142857143 1.7500000000000 0.2500000000000
8 0.0277777777778 1.7777777777778 0.2222222222222
9 0.0222222222222 1.8000000000000 0.2000000000000
10 0.0181818181818 1.8181818181818 0.1818181818182
11 0.0151515151515 1.8333333333333 0.1666666666667
12 0.0128205128205 1.8461538461538 0.1538461538462
13 0.0109890109890 1.8571428571429 0.1428571428571
14 0.0095238095238 1.8666666666667 0.1333333333333
15 0.0083333333333 1.8750000000000 0.1250000000000
16 0.0073529411765 1.8823529411765 0.1176470588235
17 0.0065359477124 1.8888888888889 0.1111111111111
18 0.0058479532164 1.8947368421053 0.1052631578947
19 0.0052631578947 1.9000000000000 0.1000000000000
20 0.0047619047619 1.9047619047619 0.0952380952381
21 0.0043290043290 1.9090909090909 0.0909090909091
22 0.0039525691700 1.9130434782609 0.0869565217391
23 0.0036231884058 1.9166666666667 0.0833333333333
24 0.0033333333333 1.9200000000000 0.0800000000000
25 0.0030769230769 1.9230769230769 0.0769230769231
26 0.0028490028490 1.9259259259259 0.0740740740741
27 0.0026455026455 1.9285714285714 0.0714285714286
28 0.0024630541872 1.9310344827586 0.0689655172414
29 0.0022988505747 1.9333333333333 0.0666666666667
30 0.0021505376344 1.9354838709677 0.0645161290323
31 0.0020161290323 1.9375000000000 0.0625000000000
32 0.0018939393939 1.9393939393939 0.0606060606061
33 0.0017825311943 1.9411764705882 0.0588235294118
34 0.0016806722689 1.9428571428571 0.0571428571429
35 0.0015873015873 1.9444444444444 0.0555555555556
36 0.0015015015015 1.9459459459459 0.0540540540541
37 0.0014224751067 1.9473684210526 0.0526315789474
38 0.0013495276653 1.9487179487179 0.0512820512821
39 0.0012820512821 1.9500000000000 0.0500000000000
40 0.0012195121951 1.9512195121951 0.0487804878049
41 0.0011614401858 1.9523809523810 0.0476190476190
42 0.0011074197121 1.9534883720930 0.0465116279070
43 0.0010570824524 1.9545454545455 0.0454545454545
44 0.0010101010101 1.9555555555556 0.0444444444444
45 0.0009661835749 1.9565217391304 0.0434782608696
46 0.0009250693802 1.9574468085106 0.0425531914894
47 0.0008865248227 1.9583333333333 0.0416666666667
48 0.0008503401361 1.9591836734694 0.0408163265306
49 0.0008163265306 1.9600000000000 0.0400000000000
50 0.0007843137255 1.9607843137255 0.0392156862745
The 2-S column was chosen as the limit seemed to be approaching 2 when the other columns were carried out further:
99996 0.0000000002000 1.9999799994000
99997 0.0000000002000 1.9999799996000
99998 0.0000000002000 1.9999799998000
99999 0.0000000002000 1.9999800000000
For n = 1 of course S = 1 and 2-S = 1. As fractions:
n: 1 2 3 4 5 6 7 8
2-S:
num: 1 2 3 4 5 6 7 8
- - - -- -- -- -- --
den: 1 3 6 10 15 21 28 36
It seems 2-S = n / ((n^2+n)/2) = 2/(n+1)
So S = 2 - 2/(n+1) verifying that S approaches 2 as n gets indefinitely larger.
|
|
Posted by Charlie
on 2022-04-14 09:59:56 |
probable solution
