Take the 2-minute tour ×

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

share|improve this question
Do you mean the sequence (X_n) with X_n = cos(n!), or are you referring to a function f(n) = cos(n!) ? Also, is this homework? –  Cam Nov 2 '10 at 21:17
No, this is not a homework, just my curiosity (from a simple exercise about limits where we had cos(n!), but the limit of it didn't matter). –  loskol Nov 2 '10 at 21:28
@Cam: those are equivalent. –  Qiaochu Yuan Nov 2 '10 at 21:59
In light of David's answer, can you mention where you found this problem, and why you think it is a true statement? –  Nate Eldredge Nov 2 '10 at 22:40
@Robert: as David's answer indicates, although it is intuitive that cos(n!) should be oscillating, actually proving it seems to require that we know much more about certain properties of pi than we actually know. –  Qiaochu Yuan Nov 3 '10 at 0:26

4 Answers 4

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.

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

share|improve this answer
I don't seem to understand your line of argument here. Could you enlighten me on this? As far as I can see you argue that if $$\frac{1}{2 \pi} = \sum_{i \geq 0} \frac{d_i}{i!}$$ such that the digits $d_i$ are bounded for some large $i$, then the limit is $1$. But what I don't see is why you consider $\frac{1}{2 \pi}$? and try to express it as $$\frac{1}{2 \pi} = \sum_{i \geq 0} \frac{d_i}{i!}$$? What if I consider $\frac{1}{\pi}$? Sorry if I sound confused. –  user17762 Nov 26 '10 at 23:51

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

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.

share|improve this answer

I tried the following approach where I assumed the limit existed and attempted to get a contradiction. However, the best I could do was to get the same answer as David above that if (in the extremely unlikely case that) the limit exists, it will be 1. I just wanted to make a record of the approach here in case someone sees any potential to this approach.

Since $x_{n+1} = \cos((n+1)n!)$, it can be written as a $(n+1)$th degree polynomial in $\cos(n!)$. Should the limit exist, it will be a fixed point of this polynomial. Using standard multiple angle formulas (found here for ex.), we can show that the limit if it exists should equal $1$.

share|improve this answer
I don't understand. The value of n changes. This argument also directly contradicts the exercise in the last part of David Speyer's answer. –  Qiaochu Yuan Nov 3 '10 at 0:02
@Qiaochu: If the limit exists, wouldn't the argument hold true for each $n$? I admit I haven't thought this through but can you elaborate on why this argument is flawed? –  Dinesh Nov 3 '10 at 0:06
@Dinesh: can you elaborate on why it isn't? It simply doesn't work. –  Qiaochu Yuan Nov 3 '10 at 0:08
@Qiaochu: I am not claiming that the limit is $1$. I am merely claiming that if one assumes that the limit exists, the limit has to be $1$ since that is the only fixed point of the $n+1$ degree polynomial. I know $n$ is changing and the polynomial changes with it but doesn't solving $x = f(x)$ give us a dirty way to find the limit of any sequence $x_1,\dots,$ such that $x_{n+1} = f(x_n)$? Am I missing something elementary? –  Dinesh Nov 3 '10 at 0:13
@Dinesh: you don't have a single f. You have a sequence of functions f_n, and as you vary n you also vary f_n. As for the second part of my objection, this argument doesn't use the fact that x_1 = cos 1. –  Qiaochu Yuan Nov 3 '10 at 0:15

Consider cos(n!) in degrees and if n>6 we can write cos(n*....*6!).6!=720=360*2.So we may assume that it turns (n*...*7) times so as a multiple of 360 the answer may be 1.I'm not sure of this answer so please check.

share|improve this answer
The asker means $n!$ radians, not $n!$ degrees. –  Casteels Aug 21 '14 at 11:54

Your Answer


