How to find such limits :

$$\lim_{m \to \infty} \lim_{n \to \infty} cos^{2m}(n! \pi x)$$

Please suggest not getting any idea how to approach such problems, will be of great help thanks.

  • 7
    $\begingroup$ That was asked and answered here. $\endgroup$
    – user64494
    Commented Sep 4, 2015 at 4:02

2 Answers 2


I saw a similar problem many years ago. The idea is that any real number can be written in the form $$ x = a_1 + \sum_{n=2}^\infty \frac{a_n}{n!} ,$$ where $a_1$ is an integer, and $a_n$ is an integer between $0$ and $n-1$. Then $$ \cos^2(n! \pi x) = \cos^2\left(\pi \left(\frac{a_{n+1}}{n+1} + O(\tfrac1{n})\right)\right) .$$ Hence the behavior of $\cos^2(n! \pi x)$ depends upon how $\frac{a_{n}}{n}$ behaves (modulo 1) as $n \to \infty$. In this manner, by the appropriate choice of $x$, you can make $$ f_x = \lim_{n\to\infty} \cos^2(n!\pi x) $$ be any number in $[0,1]$ you like, or you can even make the limit not exist.

And now you just have to worry about $\lim_{m\to\infty} f_x^m$.

It will converge to $1$ if and only if $\frac{a_{n}} n$ converges to $0$ modulo 1. For example, if $x = e$, then the limit is $1$.


Here $x$ is a constant. It can either rational or irrational.

  1. $x$ is rational.

    $x$ can be expressed as $\frac pq$ where $p$ and $q$ are natural numbers.

    $n!$ is $ n$ and as $n$ tends to infinity, $n!$ must contain $q$. Therefore $q$ cancels off and all that is left is a positive integral multiple of $\pi$.

    $\cos k\pi$ is 1 or -1. So $\cos^2 k\pi$ is 1. Therefore the answer is 1.

  2. $x$ is irrational

    Similar to the above case but the argument of the cosine function will be less than $\pi$. So the range is -1 to 1. But since $\cos^2$, the range is 0 to 1. Any number in 0 to 1 raised to infinity is 0. Therefore the answer is 0.

  • $\begingroup$ I think the answer is 1 if $x = e$. $\endgroup$ Commented Sep 4, 2015 at 3:37

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