| I am in this notes conference only about twice a year due to lack of time
(not interest). Note 1759.0 caught my attention.
My answer for (b) is ...
D ==> 2 - C - ln(4) = .036489973...
One solution for .0 ...
(a) Show D[n] is increasing by taking the first derivative of function D of a
real variable n.
Thus D'[n] = 1/n + 4/[(2n-1)^2] which is positive for all real n>=1.
Therefore D[n] must be an increasing sequence for integer n>=2
(b) To find lim D[n] as n==> infinity...
Let H(n) be the harmonic series 1 + 1/2 + 1/3 + 1/4 + ... + 1/n
It is known that for large n, and specifically in the limit
H(n) = ln(n) - C, where C is Euler's constant
[My reference for H and C above is Knuth, Volume I, Fundamental Algorithms]
(rather than C, usual symbol is lower case gamma = .5772156649... )
The series S = [ 1/3 + 1/5 + ... 1/(2n-1) ] found in .0 can be shown to
be related to H(n) as follows by manipulation of terms:
2*S = 2*H(2n-1) - H(n-1) - 2
Thus D[n] = ln(n) - 2H(2n-1) + H(n-1) + 2
= ln(n) - 2 ln (2n-1) - 2C + ln (n-1) + C + 2
n(n-1)
= ln --------- - C + 2
(2n-1)^2
lim D[n] = ln (1/4) - C + 2
= 2 - C - ln(4)
= .036489973...
Interesting result! As a check this compares within 4-5 decimal places
with a series calculation of approx D[100]=.0364842 and D[1000]=.0364919
Allen
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