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Summing up II (Posted on 2022-04-14) Difficulty: 3 of 5
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 ?

See The Solution Submitted by Ady TZIDON    
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Solution probable solution | Comment 1 of 2
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
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