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?

Question

An asymptotic behavior of       Li          −      n        ⁡  (  a  ) for   n  →  ∞Suppose   a  ,  b  ∈  (  0  ,  1  ). I&#039;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&#039;t yet found the exact exponent base or asymptotic oscillation frequency.Could you suggest a proof of (2) or further refinements of this formula?

Alec Reid · Accepted Answer

This is quite bizarre, but I believe that I can prove that your equation holds with the desired error term for any  1  &amp;gt;  a  ,  b  &amp;gt;      e          −      π        .The asymptotic that I give for       L          −      n        (  a  ) only holds when   1  &amp;gt;  a  &amp;gt;      e          −      π      By definition, for a&amp;lt;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&#039;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      =      ∞        =  0for all k&amp;lt;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  &amp;gt;  a  ,  b  &amp;gt;      e          −      π        ,and in this case it is   o  (      n          −      N        ) for any fixed capital N.

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