I found this article and read with satisfaction after I searched to confirm and streangthen my ideas.
But, I have to point out two errors that I suppose.
[1] In Page 7, regarding the equation P(k)=ln{1+1/k(k+2) }/ln2 (25),
you mentioned in the last sentence "If we make k large enough to expand the 'binominal theorem' (so that k(k+2) behaves as k^2), then P(k) ～ k^(-2) as k→∞". Its formula is not correct . What I thought at thought is the Taylor Expansion, but I found that it's actually possible when making two differece eauations and summing up those before taking its limit to the infinity:
① 2{ln(k+1)-ln k} - {ln(k+2)-ln k}. ② Then, we have Σ(k=1 to ∞) P(k)=ln2/ln2=1.
I was abe to find with your indication that this process is helpful to understand the Khnchin's Constant in his book "Continued Fractions".
[2] In Page 9, In the example for "A notable exeption", you have written the value as follows:
(k_1(e)k_2(e)...k_n(e))^(1/n) → ( 2n/3e)^(1/3)=0.62595n^(1/3). (31)
But, since e=[2; 1,2,1, 1,4,1, 1,6,1, ・・・, 1,2n,1, ・・・], the total products among the first 3n units is still 2^n*n!, then we have the following relation with the help of Stirling formula as (n!)^(1/n)= n/e+O(lon n) and then,
(k_1(e)k_2(e)...k_3n(e))^(1/3n) → ( 2n/e)^(1/3)=0.90277n^(1/3). (31)#
It has to delete the '3' in the denominator of the lefthand side of (31).
I found this article and read with satisfaction after I searched to confirm and streangthen my ideas.
But, I have to point out two errors that I suppose.
[1] In Page 7, regarding the equation P(k)=ln{1+1/k(k+2) }/ln2 (25),
you mentioned in the last sentence "If we make k large enough to expand the 'binominal theorem' (so that k(k+2) behaves as k^2), then P(k) ～ k^(-2) as k→∞". Its formula is not correct . What I thought at thought is the Taylor Expansion, but I found that it's actually possible when making two differece eauations and summing up those before taking its limit to the infinity:
① 2{ln(k+1)-ln k} - {ln(k+2)-ln k}. ② Then, we have Σ(k=1 to ∞) P(k)=ln2/ln2=1.
I was abe to find with your indication that this process is helpful to understand the Khnchin's Constant in his book "Continued Fractions".
[2] In Page 9, In the example for "A notable exeption", you have written the value as follows:
(k_1(e)k_2(e)...k_n(e))^(1/n) → ( 2n/3e)^(1/3)=0.62595n^(1/3). (31)
But, since e=[2; 1,2,1, 1,4,1, 1,6,1, ・・・, 1,2n,1, ・・・], the total products among the first 3n units is still 2^n*n!, then we have the following relation with the help of Stirling formula as (n!)^(1/n)= n/e+O(lon n) and then,
(k_1(e)k_2(e)...k_3n(e))^(1/3n) → ( 2n/e)^(1/3)=0.90277n^(1/3). (31)#
It has to delete the '3' in the denominator of the lefthand side of (31).