functions – Does Mathematica have a problem with sums involving Stirling numbers of the second kind?

This is a comment and not an answer, but I wanted to upload a picture.

You say that it’s a very simple summation, but does it have a known closed form? If it does, then it seems Mathematica is not programmed with the tools to achieve it through Sum. If you post the closed form, someone here might be better able to help you.

However, looking at the length of the numerator and denominator seems to suggest that they grow continuously.

tbl = Table(
    IntegerLength /@ NumeratorDenominator(
      Sum((StirlingS2(k - 1, 4) + StirlingS2(k, 4))/6^k, {k, n})), 
    {n, 1000})(Transpose);

Plot of the numerator and denominator of the sum.

The length of the numerator is plotted in blue and the denominator in yellow. I realize that this is only the first 1000 terms and a far cry from infinity, but is there a reason to expect the length of the numerator or denominator to plateau at some point? It seems to me like the size of the representation of that number will continue to grow practically forever.

A linear fit to the data suggests that the numerator will grow to 7780.04 digits by the time you sum to 10,000, and

    Sum((StirlingS2(k - 1, 4) + StirlingS2(k, 4))/6^k, {k, 10^4})
(* 7780 *)

Seems to confirm that. As @Bill said, the sum seems to “converge” fairly quickly on any scale that we would normally care about, but the exact representation of that value seems to grow continuously. If my calculation is correct, and depending on your computer’s RAM, you would probably lose the ability to even store the number in memory by the time $n = 10^{11}$ or so.