Proving Limit of sin (n!*R*π) for Rational R

[happyg1]

Hi,
Here is my dilemma: I am to prove that sin (n!*R*pi) has a limit, where R is a rational number. I rewrite R as a/b and I can see that whenever n=b, every subsequent term will be zero. I have tried to write this out using the definition of a limit, but I can't seem to break it down. I have been looking at these problems for a long time and I am blocked on this one.
Thanks in advance,
CC

 

[Physics Monkey]

For a sequence $$a_n$$, we say that $$a_n \rightarrow L$$ if for every $$\epsilon > 0$$ we can find an $$N(\epsilon)$$ such that $$| a_n - L | < \epsilon$$ for $$n > N(\epsilon)$$.

You have guessed correctly that the limit is 0. Now I give you an $$\epsilon$$ and ask you to tell me where I should start looking so that the terms of the sequence are always closer than $$\epsilon$$ to the limit. Tell me where to look by giving me N. You've already pointed out that the terms of the sequence equal the limit beyond a certain point ...

 

[happyg1]

hey,
I think I got it. I wasn't including my b in the expansion of the n! as the spot where the sequence converges. I couln't relate the epsilon to the b or the n. It's just the algebra. That's what was giving me the headache. I had been doing the ones that are all polynomials and my brain was fried.

CC