A sequence S(n) is defined as:

S(1) = 0, and:
S(n) = n*S(n-1)+2*n-1 with n > 1

Determine:

Limit    S(n)/n!
n → ∞

Private Sub Form_Load()
 Text1.Text = ""
 s = 0: n = 1: f = 1
 For n = 2 To 30
   s = n * s + 2 * n - 1
   f = f * n
   Text1.Text = Text1.Text & Str(n) & Str(s / f) & Chr(13) & Chr(10)
 Next
 
End Sub

finds

 2 1.5
 3 2.33333333333333
 4 2.625
 5 2.7
 6 2.71527777777778
 7 2.71785714285714
 8 2.71822916666667
 9 2.71827601410935
 10 2.71828125
 11 2.71828177609428
 12 2.71828182411082
 13 2.71828182812558
 14 2.71828182843529
 15 2.71828182845747
 16 2.71828182845895
 17 2.71828182845904
 18 2.71828182845905
 19 2.71828182845905
 20 2.71828182845905
 21 2.71828182845905
 22 2.71828182845905
 23 2.71828182845905
 24 2.71828182845905
 25 2.71828182845905
 26 2.71828182845905
 27 2.71828182845905
 28 2.71828182845905
 29 2.71828182845905
 30 2.71828182845905

At n=18 S(n)/n! has reached the value of e to the accuracy of the double-precision numbers used in the program.

Posted by Charlie on 2014-04-21 13:46:49