Solve or Reduce don't work with Integer functions

Question

I had been trying to limit for some c++ programs, the bigest whole number they can handle is

lim = 18446744073709551615

and the different functions are:

tfn[k_] := Total[Table[n!, {n, k}]]

or

spd2[n_] := Total[Table[2^i, {i, n}]]

the only way I found to know the limit k that do not overflow is

klimtfn = For[i = 1, 1 < 100, i++, If[tfn[i] > lim, Break[]]]; i - 1

or

klimspd2 = For[i = 1, 1 < 100, i++, If[spd2[i] > lim, Break[]]]; i - 1

for both cases I found correct answers, my concern is I have used forand this is like unfair, surely I must learn something better otherwise. Thanks for advice.

Answer 1

Mathematica can do the sums symbolically, after which you can use FindRoot (actually, the symbolic sum isn't necessary, as can be seen by replacing = with := in the following). First example:

f[k_] = Sum[n!, {n, k}]

-1 - Subfactorial[-1] + (-1)^(1 + k) Gamma[2 + k] Subfactorial[-2 - k]

Using FindRoot:

k1 = k /. FindRoot[f[k]== 18446744073709551615, {k, 40},WorkingPrecision->16]

20.65081840115899

Check:

Sum[n!, {n, Floor[k1]}] < 18446744073709551615 < Sum[n!, {n, Floor[k1]+1}]

True

For your second example:

g[k_] = Sum[2^i, {i, k}]

2 (-1 + 2^k)

Using FindRoot:

k2 = k /. FindRoot[g[k] == 18446744073709551615, {k, 100}]

63.

Check:

Sum[2^i, {i, Floor[k2]}] < 18446744073709551615 < Sum[2^i, {i, Floor[k2]+1}]

True