Is there a limit of $\cos(n!)$?

Question

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

Answer 1

I suspect that you have miscopied the problem, or your source has an error. I will argue that, while your statement is extremely likely to be true, we probably do not know enough to prove it.

Here is why. Suppose, hypothetically, that $$\frac{1}{2 \pi} = \sum_{i \geq 0} \frac{d_i}{i!}$$ where all the $d_i$ are integers and, for $i$ sufficiently large, each $d_i$ is either $0$ or $1$. This is extraordinarily unlikely. However, if we knew how to prove that this didn't happen, we would probably know enough to show that every digit occurs infinitely often in the decimal expansion of $\pi$, and I believe that is an open problem. So, while this scenario is almost surely false, I don't think we know enough to prove that it is false.

Now, in this scenario, $$\cos n! = \cos (n! 2 \pi \sum_{i \geq 0} \frac{d_i}{i!}) = \cos (2 \pi N + 2 \pi \sum_{i \geq n+1} \frac{d_i }{(n+1)(n+2) \cdots i})$$ where $N$ is a very large integer. By the periodicity of $\cos$, this is just $$\cos (2 \pi\sum_{i \geq n+1} \frac{d_i }{(n+1)(n+2) \cdots i}).$$

Now, once $n$ is large enough that $d_i$ is $0$ or $1$ for $i$ larger than $n$, we have $$\sum_{i \geq n+1} \frac{d_i n!}{i!} \leq \frac{1}{n+1} + \frac{1}{(n+1)^2} + \frac{1}{(n+1)^3} + \cdots = \frac{1}{n}.$$ So, in our unlikely scenario, $\lim_{n \to \infty} \sum_{i \geq n+1} \frac{d_i n!}{i!} =0$ and $\lim_{n \to \infty} \cos n! = 1$.

If I may try to extract some basic lessons here, questions about trig functions evaluated at integers almost always turn into questions about $\pi$. Often they can be resolved simply by noting that $\pi$ is irrational; occasionally it is important to show that $\pi$ is not well approximated by rationals and sometimes, as in this example, they turn into claims about $\pi$ which are beyond current knowledge. The key point is to work out what they are saying about $\pi$, and then find out whether that fact is known.

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Exercise: Show that, for any $a$ between $-1$ and $1$, there is a $\theta$ such that $\lim_{n \to \infty} \cos (n! \theta)$ is $a$.

Answer 2

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:

<blockquote>
  <p>eps = .00001<br>
  for i in xrange(2,800000,2):<br>
  ...     if i*math.pi-math.floor(i*math.pi) &lt; eps: print i<br>
  ...<br>
  431230<br>
  530762<br>
  630294  </p>

  <p>math.cos(math.floor(431230*math.pi)) =
  0.99999999997167566<br>
  So any n! for n > 431230*pi will be at least this close, or closer, to 1  </p>
</blockquote>

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.