Getting wrong limit with Bessel

Question

Bug introduced in 7.0 or earlier and fixed in 11.3

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I computed a following limit (related to the asymptotic expansion of the sequence A000009 - number of partitions of n into distinct parts)

Expand[Limit[(((π BesselI[1, (Sqrt[1/24 + n] π)/Sqrt[3]])/Sqrt[1 + 24 n])/
             (E^((Sqrt[n] π)/Sqrt[3])/(4 3^(1/4) n^(3/4))) -
             (1 + (-((3 Sqrt[3])/(8 π)) + π/(48 Sqrt[3]))/Sqrt[n] +
              (-(5/128) - 45/(128 π^2) + π^2/13824)/n))*n^(3/2), n -> Infinity]]

Mathematica wrong output (in all versions from 7 to 11) is

(* (35 Sqrt[3])/(2048 π) - (35 π)/(36864 Sqrt[3]) + π^3/(1990656 Sqrt[3]) *)

This expression is equal to

(* 0.007709031447942101952246679 *)

But the correct result is

(* -((315 Sqrt[3])/(1024 π^3)) + (35 Sqrt[3])/(2048 π) -
   (35 π)/(36864 Sqrt[3]) + π^3/(1990656 Sqrt[3]) *)

This is equal to

(* -0.009474863397773687074232327 *)

My question is: why is the term

(* -((315 Sqrt[3])/(1024 π^3)) *)

missing in the Mathematica result? This is a bug!

Interesting is that numerically Mathematica evaluates a following expression correctly:

Table[N[(((π BesselI[1, (Sqrt[1/24 + n] π)/Sqrt[3]])/Sqrt[1 + 24 n])/
         (E^((Sqrt[n] π)/Sqrt[3])/(4 3^(1/4) n^(3/4))) -
         (1 + (-((3 Sqrt[3])/(8 π)) + π/(48 Sqrt[3]))/Sqrt[n] +
          (-(5/128) - 45/(128 π^2) + π^2/13824)/n))*n^(3/2), 10],
      {n, 1000000, 10000000, 1000000}]

(* {-0.009484688338, -0.009481807906, -0.009480532562, -0.009479772520, -0.009479253935, -0.009478871178, -0.009478573728, -0.009478333979, -0.009478135403, -0.009477967422} *)

Maple evaluates this limit correctly. Here is the Maple code:

expand(simplify(limit((Pi * BesselI(1, Pi * sqrt((n+1/24) * (1/3)))*(4 * 3^(1/4) * n^(3/4))/(sqrt(24 * n+1)*exp(Pi * sqrt((1/3) * n)))-1-(Pi/(48 * sqrt(3))-3 * sqrt(3)/(8 * Pi))/sqrt(n)-((1/13824) * Pi^2-5/128-45/(128 * Pi^2))/n) * n^(3/2), n = infinity))); evalf(%, 60);

Answer 1

I think it is something like a bug of Mathematica: It seems that Mathematica uses Series in order to calculate the Limit. When you take Series up to a expansion order that is high enough, you get the right answer. Mathematica doesn't take into account that the summation of Sqrt[1/n], n-> Infinity terms is not limited. When you take also the (1/n)3/2 term you get the right result. Look:

f[n_] = 
  (((π BesselI[1, (Sqrt[1/24 + n] π)/Sqrt[3]])/
    Sqrt[1 + 
      24 n])/(E^((Sqrt[n] π)/
       Sqrt[3])/(4 3^(1/4) n^(3/
         4))) - (1 + (-((3 Sqrt[3])/(8 π)) + π/(48 Sqrt[
          3]))/Sqrt[
     n] + (-(5/128) - 45/(128 π^2) + π^2/13824)/n))*
 n^(3/2) // Expand

Table[Series[f[n], {n, Infinity, i}], {i, 0, 3}]

It's to long to show here.

Table[Limit[Series[f[n], {n, Infinity, i}] // Normal, n -> Infinity], {i, 0, 6}]

{0, 0, (102060 - 1890 π^2 + π^4)/(
   1990656 Sqrt[3] π), (-1837080 + 102060 π^2 - 
   1890 π^4 + π^6)/(
  1990656 Sqrt[3] π^3), (-1837080 + 102060 π^2 - 
  1890 π^4 + π^6)/(
  1990656 Sqrt[3] π^3), (-1837080 + 102060 π^2 - 
  1890 π^4 + π^6)/(
   1990656 Sqrt[3] π^3), (-1837080 + 102060 π^2 - 
    1890 π^4 + π^6)/(1990656 Sqrt[3] π^3)}

Table[Limit[Series[f[n], {n, Infinity, i}] // Normal, n -> Infinity], {i, 0, 6}] // N

{0., 0., 0.00770903, -0.00947486, -0.00947486, -0.00947486, -0.00947486}