# Problem computing a limit

I have the following function in Mathematica:

$$\frac{x! \sum _{i=1}^j (-i+j+1) S_{j+1}^{(-i+j+2)} x^{-i+j+1}}{(x-j)! \left((1-x)_j\right){}^2}$$

Defined as:

a[n_, m_] := (n - m)*StirlingS1[n + 1, n + 1 - m]
p[x_, j_] := Sum[a[j, i - 1]*x^(j - i + 1), {i, 1, j}]
m[x_, j_] := (Factorial[x]/
Factorial[x - j])*(p[x, j])/(Product[(x - i)^2, {i, 1, j}])


When I try to compute the limit of m[x, j]/j when $$x$$ approaches infinity, it takes too long to provide a solution:

Limit[m[x, j]/j, x -> Infinity]

How can I get the solution of this limit in a reasonable amount of time?

Sum[] is difficult to handle in general, and it seems to be why Limit[] is so slow. In this case, it's a polynomial and its dominant term will determine the limit. If we replace the sum by this term, Limit[] returns in a reasonable amount of time. One can simply replace it by inspection (it's j x^j StirlingS1[1 + j, 1 + j]) or use the following ad hoc utility:

dominantTerm[HoldPattern[Sum[f_, {i_, a_, b_}]], var_] :=
f /. Last@Simplify@
Maximize[{First@Cases[f, var^p_ :> p, Infinity], a <= i <= b}, i];

Assuming[j > 0 && j \[Element] Integers,
FullSimplify@
Limit[m[x, j]/j // FunctionExpand //
ReplaceAll[s_Sum :> dominantTerm[s, x]], x -> Infinity]
] // AbsoluteTiming

(*  {31.6444, 1}  *)


We can speed things up by using an asymptotic approximation:

Assuming[j > 0 && j \[Element] Integers,
With[{asym =
Normal@Series[
m[x, j]/j // FunctionExpand //
ReplaceAll[s_Sum :> dominantTerm[s, x]], {x, Infinity, 0},
Assumptions -> j > 0 && j \[Element] Integers && x > j]},
FullSimplify@Limit[asym, x -> Infinity]
]] // AbsoluteTiming

(*  {2.31803, 1}  *)


The limit is 1.

The problems seems that for different $$j$$ you can get 1/0 division depending on what $$x$$ value is.

So without having specific value of $$j$$ might not be possible. But this below seems to show the limit is $$1$$.

Limit[(m[x, j]/j) /. j -> 10, {x -> Infinity}]
(*1*)

Limit[(m[x, j]/j) /. j -> 20, {x -> Infinity}]
(*1*)

Limit[(m[x, j]/j) /. j -> 100, {x -> Infinity}]
(*1*)

• Is there any way to prove that the limit is $1$ for all positive $j$? Commented Jul 10 at 14:19