Prove that a function containing $n!$ is decreasing

Question

Taking the derivative of the function $f(n)=n!$ with $n \in \mathbb{N}$ is not possible. But is there some alternative way to understand if a function with $n!$ is decreasing (besides the definition of decreasing, i.e. proving that $f(n+1)<f(n)$ directly, which is hard to do in some cases).

For example: suppose that I need to use Leibniz rule of convergence for this series. $$\sum_{n \geq 1} (-1)^n \frac{\mathrm{sin}(n!)}{n^5}$$ Firstly I must prove that $ \frac{\mathrm{sin}(n!)}{n^5}$ is decreasing, but how can I do without taking the derivative?

Answer 1

HINT:

$$|\sin(x)|\le 1$$ for any $x$. Thus, with $x=n!$, we have

$$|\sin(n!)|\le 1$$

Answer 2

The sequence $\frac{sin(n!)}{n^5}$ is not decreasing: at $n=11$ and $n=12$, its value goes from $1.255732774 \times 10^{-6}$ to $2.582485596 \times 10^{-6}$.

You should use Dr. MV's hint, though, and a different test for convergence. The $n^5$ is very helpful for you: note in particular that $5 > 2$.