TITLE: Is there a limit of $\cos(n!)$?
QUESTION [28 upvotes]: I encountered a problem today to prove that $(X_n)$ with $X_n = \cos(n!)$ does not have a limit (when $n$ approaches infinity). I have no idea how to do it formally. Could someone help? The simpler the proof (by that I mean less complex theorems are used) the better. Thanks

REPLY [2 votes]: Above posters seem to be right, I thought I had this solved for the limit = 1, but realized I was wrong.
If its of any use, here is my faulty reasoning. 


As we get more and more factors we will only get closer and closer to a factor that is a multiple of pi.  We know that n! is even for any n >=2, so we get a abritarily closer to 2*pi|n! as n approaches infinity.
    Sorry of the code is out of form, but:

eps = .00001
      for i in xrange(2,800000,2):
      ...     if i*math.pi-math.floor(i*math.pi) < eps: print i
      ...
      431230
      530762
      630294  
math.cos(math.floor(431230*math.pi)) =
      0.99999999997167566
      So any n! for n > 431230*pi will be at least this close, or closer, to 1  



The problem is that product of the factors NOT close to pi is going to grow faster than that epsilon shrinks.  And obviously no integer is going to be an exact multiple of pi (as it is irrational).  Therefore, I am led to agree with other posters that this limit is unlikely.
PS - I know this isn't rigorous by any means, but was hoping maybe better minds would have some insightful comments.