1 / n!

[killo]

Just a simple question, I'm using the series 1 + 1/1! + 1/2! + 1/3! ... + 1/n! to compute e, the first 3 terms: 1, 1/1!, 1/2! come out as 1, 1, 0.5, but it seems all the other terms contain recurring decimals, like 1/3! = 1/6 = 0.166666.....
I was wondering if there were any terms after n=2 for 1/n! where the result does not come out as a recurring decimal, I thought it's likely there isn't, but is there a proof?

[jemidiah]

The Repeating decimal Wikipedia page gives the period of numbers of this form. You can factor out the 2 and 5 terms of n!'s prime factorization, apply the linked result to say that number's reciprocal is non-terminating for n>2 (since then it must contain a 3 at the least), note that the factored term will terminate, and finally apply the fact (derivable from the standard multiplication algorithm) that the product of a non-terminating and a terminating decimal is a non-terminating decimal. Your result follows

I'd enjoy a cleaner proof if anyone has one.


Edit: I got to thinking, the product of a terminating and a non-terminating number isn't necessarily non-terminating. However it's sufficient in this case to show that multiplying a non-terminating number by either 1/2 or 1/5 yields a non-terminating number (proceeding inductively). Since 1/5 = 0.2, consider doubling a non-terminating number. A proof by cases (that's a bit tedious and I don't wanna go through) should give the required result.

[killo]

Thanks, I think I understand what you're saying, basically when n>2 for n! all terms will contain a "x3" because 3! = 1x2x3 and 4! = 1x2x3x4 and so on, and because they all contain "x3" then 1/n! will always be a non terminating decimal for n>2.

[jemidiah]

Yeah that's the basic idea; the rest is making a rigorous proof for it.