"An asymptotic behavior of Li−n(a) for n-> inf Suppose a,b in (0,1). I'm interested in comparison of an asymptotic behavior of Li−n(a) and Li−n(b) for n-> inf. Such functions exhibit approximately factorial-like (faster than exponential) growth rate. The particular case Li−n(12) for n>=1 gives (up to a coefficient) a combinatorial sequence called Fubini numbers or orderded Bell numbers[1][2][3] (number of outcomes of a horse race provided that ties are possible).

mydaruma25 2022-09-17 Answered
An asymptotic behavior of Li n ( a ) for n
Suppose a , b ( 0 , 1 ). I'm interested in comparison of an asymptotic behavior of Li n ( a ) and Li n ( b ) for n .
Such functions exhibit approximately factorial-like (faster than exponential) growth rate. The particular case Li n ( 1 2 ) for n 1 gives (up to a coefficient) a combinatorial sequence called Fubini numbers or orderded Bell numbers[1][2][3] (number of outcomes of a horse race provided that ties are possible). This sequence is known to have the following asymptotic behavior:
(1) Li n ( 1 2 ) n ! ln n + 1 2 .
After some numerical exprerimentation I conjectured the following behavior:
(2) ln ( Li n ( a ) Li n ( b ) ) = ( n + 1 ) ln ( ln b ln a ) + o ( n N )
for arbitrarily large N (so, the remainder term decays faster than any negative power of n). It looks like the remainder term is oscillating with exponentially decreasing amplitude, but I haven't yet found the exact exponent base or asymptotic oscillation frequency.
Could you suggest a proof of (2) or further refinements of this formula?
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Answers (1)

Alec Reid
Answered 2022-09-18 Author has 9 answers
This is quite bizarre, but I believe that I can prove that your equation holds with the desired error term for any
1 > a , b > e π .
The asymptotic that I give for L n ( a ) only holds when 1 > a > e π
By definition, for a<1
Li n ( a ) = k = 1 k n a k ,
and we can approximate this sum by the integral
0 x n a x d x = 0 x n e x log ( 1 a ) d x .
Letting x = u log 1 / a the integral becomes
1 ( log ( 1 a ) ) n + 1 0 u n e u d x = Γ ( n + 1 ) ( log ( 1 a ) ) n + 1 ,
and so
Li n ( a ) Li n ( b ) ( log ( 1 / b ) log ( 1 / a ) ) n + 1 .
Let's prove this with an exact error term by making the first step precise. Let f a ( x ) = x n a x / Then since
d k d x k f a ( x ) | x = 0 = d k d x k f a ( x ) | x = = 0
for all k<n, Euler-Maclaurin summation up to p = n 2 for even n implies that
| k = 0 k n a k 0 x n a x d x | | R | 2 ζ ( n ) ( 2 π ) n ( 0 | f ( n ) ( x ) | d x ) 2 ζ ( n ) ( 2 π ) n ( 2 n Γ ( n + 1 ) log ( 1 / a ) ) = O ( 1 π n Γ ( n + 1 ) log ( 1 / a ) ) .
Thus,
Li n ( a ) Li n ( b ) = Γ ( n + 1 ) ( log ( 1 a ) ) n + 1 + O ( 1 π n Γ ( n + 1 ) log ( 1 / a ) ) Γ ( n + 1 ) ( log ( 1 b ) ) n + 1 + O ( 1 π n Γ ( n + 1 ) log ( 1 / b ) ) = ( log ( 1 / b ) log ( 1 / a ) ) n + 1 ( 1 + O ( π n log ( 1 / a ) n ) 1 + O ( π n log ( 1 / b ) n ) ) ,
and so
Li n ( a ) Li n ( b ) = ( log ( 1 / b ) log ( 1 / a ) ) n + 1 ( 1 + O ( π n log ( 1 min ( a , b ) ) n ) ) ,
and using the fact that l log ( 1 + x ) = x + O ( x 2 ) , this implies implies that
log ( Li n ( a ) Li n ( b ) ) = ( n + 1 ) log ( log ( 1 / b ) log ( 1 / a ) ) + O ( π n log ( 1 min ( a , b ) ) n ) .
Now, this error term is nontrivial only when
1 > a , b > e π ,
and in this case it is o ( n N ) for any fixed capital N.

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