Try performing the long division necessary to get the digits.
One you might notice is that there are only a finite set of possible remainders, and any particular remainder is always followed by the same next remainder. It's easy to see why; at each step you add a 0 and divide again, so same remainder means same next step.
For example, I'll do it with 7 (with somewhat modified notation; I hope it's clear):
7 | 1 -> 0 R 1
7 | 10 -> 0.1 R 3
7 | 30 -> 0.14 R 2
7 | 20 -> 0.142 R 6
7 | 60 -> 0.1428 R 4
7 | 40 -> 0.14285 R 5
7 | 50 -> 0.142857 R 1
7 | 10 -> 0.1428571 R 3
...
The sequence of remainders will repeat forever (1, 3, 2, 6, 4, 5), which will forever generate the sequence of digits (142857).
There are only 7 possible remainders when you divide by 7; 0, 1, 2, 3, 4, 5, and 6. The last 6 of them are in the above cycle, so any time you get one of those remainders, the following digits will follow that cycle. The only other possibility is that 7 evenly divides the number in question (such as 14 / 7), where there is only one remainder in the cycle (0) and the sequence of digits is simply an infinite string of 0's.
Every other number behaves similarly, except 7 is somewhat special by having one large cycle of non zeroes. For example, with the number 4:
Starting with a remainder of 0, 00 / 4 = 0 R 0, so remainder 0 progresses to remainder 0 yielding digit 0
Starting with a remainder of 1, 10 / 4 = 2 R 2, so remainder 1 progresses to remainder 2 yielding digit 2
Starting with a remainder of 2, 20 / 4 = 5 R 0, so remainder 2 progresses to remainder 0 yielding digit 5
Starting with a remainder of 3, 30 / 4 = 7 R 2, so remainer 3 progresses to remainder 2 yielding digit 7
So, if you were computing something like 3/4, you get a remainder of 3, which yields remainder 2 then an infinite sequence of 0's, which gives the digits 7, 5, 0, 0, 0...
i.e. 3 / 4 = 0.75000000...
Other numbers have different cycle structures. 6 has several cycles, which can be observed in its somewhat interesting diversity of expansions:
0/6 = 0.00000...
1/6 = 0.16666...
2/6 = 0.33333...
3/6 = 0.50000...
4/6 = 0.66666...
5/6 = 0.83333...
The number 11 has a whole buch of two cycles, yielding the familiar expansions:
1/11 = 0.090909...
2/11 = 0.181818...
3/11 = 0.272727...
et cetera
The number 13 is more similar to 7; it yields two separate 6-cycles, but not a 12 cycle (of course, it also has the trivial all 0's cycle). The two corresponding patterns are:
(076923) and (153846)
And any decimal expansion of a fraction with denominator 13 will end in a repeating sequence of one of those cycles of digits. For example:
4/13 = 0.3(076923)07692...
I don't know if any larger numbers have a complete cycle like 7 does; it has been a long time since I played with this, and I don't think I went past 13.
