Can cos(n!) in degrees tend to one if n>6?

Question

does cos(n!) in degrees tend to 1. consider cos(n!)=cos(n*...*6!),6!=720=360*2.So this is like rotating on the plane n*...7*2 times so cos(n!)=1,When n>6 .Does this proof hold even when n tends to infinity.please give a reply which suits the mind of a beginner.

Answer 1

Indeed $\cos (n!) = 1$ for $n \geq 6$ since $n! \equiv 0 \pmod{360}$. But $n$ must be a finite number, $\cos \infty$ makes no sense, but we can say that $\displaystyle\lim_{n\to\infty} \cos (n!) = 1$:

We say that the sequence $\{a_n\}$ tends to the finite limit $l$ as $n$ goes to infinity if:

$\forall \varepsilon >0\ \exists n_0 > 0$ so that if $n > n_0$ then $|a_n - l| < \varepsilon$.

Taking $l = 1$ and $n_0 = 5$ then regardless of how $\varepsilon$ is, the definition holds, so indeed, $$\lim_{n\to\infty} \cos (n!) = 1$$

Answer 2

The argument you gave is entirely correct. In $n \ge 6$, then $$ $\cos(n!) = \cos\big( n \times (n-1) \times (n-2)\times\ldots\times 8 \times 7\times 720\big) = \cos(360) = 1 $$ So, for all $n$ greater than $6$, we have $\cos(n!)=1$, and this certainly implies that $\cos(n!)$ tends to $1$ as $n$ tends to infinity. The mathematical way of saying this is $$ \lim_{n \to \infty} \cos(n!) = 1 $$

But, as others have said, it doesn't make sense to simply substitute $n=\infty$ in the expression $\cos(n!)$, because $\cos(x)$ only makes sense when $x$ is a number, and $\infty$ is not a number (or, not one of the right type, anyway).

Answer 3

Using Riemann zeta function,infinite factorial can be regularized $$\prod_{n=1}^\infty\lambda_n=\exp(-\zeta'_{\lambda}(0))\\\zeta_\lambda(s)=\sum_{n=1}^\infty \lambda_n^{-s}\\\zeta_\lambda'(s)=-\sum_{n=1}^\infty\lambda_n^{-s}\log(\lambda_n)$$ Plugging in $\lambda_n=n$ we get that $$\prod_{n=1}^\infty n=\infty!=\sqrt{2\pi}$$ NOTE: $\zeta'(0)=-\frac{1}{2}\ln({2\pi})$ which can be derived directly from the Wallis formula

So $\cos(\infty!)=\cos(\sqrt{2\pi})\approx0.99904$