Liouville's number is transcendental. Nonetheless one can still make an infinite series using it to produce an integer! I came across the procedure in a previous message. Can anyone figure out how I could have done it? Here is an example.

The infinite series is

(313222/285 + (280 l)/171)*10^5 - 7/4275*10^-16 - 
  Sum[28/171*10^(-n! + 6), {n, 5, Infinity}] == 109920468.

A proviso is that this is not a type of series you want Mathematica to evaluate in total. That is because

 N[(313222/285 + (280 l)/171)*10^5 - 7/4275*10^-16 - Sum[28/171*10^(-n! + 6), {n, 5, 5}]
   -
  ((313222/285 + (280 l)/171)*10^5 - 7/4275*10^-16 - Sum[28/171*10^(-n! + 6), {n, 5, Infinity}])]

gives 0. instead of a small residue.

First make a good enough approximation to Liouville's number such that it produces no round-off error in the computation:

l = NSum[10^-n!, {n, 1, 10}, WorkingPrecision -> 8!];

Truncate your infinite series to a finite number of partial sums so that Matheamtica can do arithmetic with it:

f[x_] := (313222/285 + (280 l)/171)*10^5 - 7/4275*10^-16 - 
   Sum[28/171*10^(-n! + 6), {n, 5, x}];

Finally, approximate your answers:

  N[109920468 - Table[f[x], {x, 5, 7}]]

(* {-1.637426900584795*10^-715, -1.637426900584795*10^-5035,  -2.*10^-40315}*)

I conjectured that one can find such infinite series equaling every integer.

There might be another