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It's e-asy (Posted on 2008-02-19) Difficulty: 3 of 5
Outline a method for calculating Sk = Σ nk/n! for n=1 to infinity and compute the first few terms of Sk for natural numbers k.

See The Solution Submitted by FrankM    
Rating: 2.0000 (2 votes)

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Solution computer solution | Comment 1 of 7

Hmmm....

Calculating n^k is easy enough by multiplying n by itself k times, or you could take the antilog of k*log(n). And n! is easier, as it builds on (n-1)!, so you don't even have to start from scratch with each term.

The numbers get big as n gets large, but the factorials increase so greatly compared to the powers that only a few terms are needed.

DEFDBL A-Z

FOR k = 1 TO 10
  PRINT "k ="; k
  t = 0
  term = 1: f = 1
  FOR n = 1 TO 44
   f = f * n
   term = n ^ k / f
   t = t + term
   PRINT USING "## ####.######### ####.########"; n; term; t
  NEXT
  PRINT k; t;
  IF k = 1 THEN save = t:  ELSE PRINT INT(t / save + .5);
  PRINT
  DO: LOOP UNTIL INKEY$ > ""
NEXT

Some results:

k = 1
 n      nth term    partial total
 1    1.000000000    1.00000000
 2    1.000000000    2.00000000
 3    0.500000000    2.50000000
 4    0.166666667    2.66666667
 5    0.041666667    2.70833333
 6    0.008333333    2.71666667
 7    0.001388889    2.71805556
 8    0.000198413    2.71825397
 9    0.000024802    2.71827877
10    0.000002756    2.71828153
11    0.000000276    2.71828180
12    0.000000025    2.71828183
13    0.000000002    2.71828183
14    0.000000000    2.71828183

So, for k=1, it seems to converge to e.

k = 2
 1    1.000000000    1.00000000
 2    2.000000000    3.00000000
 3    1.500000000    4.50000000
 4    0.666666667    5.16666667
 5    0.208333333    5.37500000
 6    0.050000000    5.42500000
 7    0.009722222    5.43472222
 8    0.001587302    5.43630952
 9    0.000223214    5.43653274
10    0.000027557    5.43656030
11    0.000003031    5.43656333
12    0.000000301    5.43656363
13    0.000000027    5.43656365
14    0.000000002    5.43656366
15    0.000000000    5.43656366

So, for k=2, it seems to converge to 2e.

k = 3
 1    1.000000000    1.00000000
 2    4.000000000    5.00000000
 3    4.500000000    9.50000000
 4    2.666666667   12.16666667
 5    1.041666667   13.20833333
 6    0.300000000   13.50833333
 7    0.068055556   13.57638889
 8    0.012698413   13.58908730
 9    0.002008929   13.59109623
10    0.000275573   13.59137180
11    0.000033344   13.59140515
12    0.000003608   13.59140876
13    0.000000353   13.59140911
14    0.000000031   13.59140914
15    0.000000003   13.59140914
16    0.000000000   13.59140914

At k=3, we now have 5e.

k = 4
 1    1.000000000    1.00000000
 2    8.000000000    9.00000000
 3   13.500000000   22.50000000
 4   10.666666667   33.16666667
 5    5.208333333   38.37500000
 6    1.800000000   40.17500000
 7    0.476388889   40.65138889
 8    0.101587302   40.75297619
 9    0.018080357   40.77105655
10    0.002755732   40.77381228
11    0.000366788   40.77417907
12    0.000043290   40.77422236
13    0.000004587   40.77422694
14    0.000000441   40.77422738
15    0.000000039   40.77422742
16    0.000000003   40.77422743
17    0.000000000   40.77422743

Now, at k=4, it seems we have 15e. There go our hopes of getting some sequence of Fibonacci multiples of e.

k = 5
 1    1.000000000    1.00000000
 2   16.000000000   17.00000000
 3   40.500000000   57.50000000
 4   42.666666667  100.16666667
 5   26.041666667  126.20833333
 6   10.800000000  137.00833333
 7    3.334722222  140.34305556
 8    0.812698413  141.15575397
 9    0.162723214  141.31847718
10    0.027557319  141.34603450
11    0.004034667  141.35006917
12    0.000519481  141.35058865
13    0.000059626  141.35064828
14    0.000006169  141.35065444
15    0.000000581  141.35065503
16    0.000000050  141.35065508
17    0.000000004  141.35065508
18    0.000000000  141.35065508

The total is 52e.

Cutting to the chase, commenting out the individual terms, and getting the final results of each k:

 k       sum            multiple of e
 1  2.718281828459046       1
 2  5.436563656918090       2
 3  13.59140914229523       5
 4  40.77422742688568      15
 5  141.3506550798703      52
 6  551.8112111771861     203
 7  2383.933163558582     877
 8  11253.68676982045    4140
 9  57483.50582642343   21147
10  315252.7350555378  115975

Looking up the sequence 1, 2, 5, 15, 52, 203, 877 in Sloane's On-line Encyclopedia of Integer sequences leads us to "Bell or exponential numbers: ways of placing n labeled balls into n indistinguishable boxes."  So if you have a way of computing the Bell numbers, you can multiply e by the Bell numbers to get these sums.

But the ways of getting the Bell numbers look as complicated, or more so, than the computations in this program. Of course, you could start with a list of the Bell numbers and just multiply by e.

In fact, this could be one way of calculating the Bell numbers:

1  2.718281828459046                1
2  5.43656365691809                 2
3  13.59140914229523                5
4  40.77422742688568               15
5  141.3506550798703               52
6  551.8112111771861              203
7  2383.933163558582              877
8  11253.68676982045             4140
9  57483.50582642343            21147
10  315252.7350555378          115975
11  1844544.500337454          678570
12  11453744.15754954         4213597
13  75145370.75508085        27644437
14  518918158.0577521       190899322
15  3759271082.385662      1382958545
16  28487979957.85786     10480142147
17  225250069805.8378     82864869804
18  1854076987795.393    682076806159
19  15855037146092.58   5832742205057
20  140600839423552    51724158235372
21  1290829992142583  474869816156751
22  1.225052349785174D+16  4506715738447322

Only the limitations on the accuracy of this particular programming language make the last Bell number approximation off by 1 in the units position.


  Posted by Charlie on 2008-02-19 12:42:02
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