how to show that $\sin(\frac{1}{n})$ is a decreasing function

Using the Prosthaphaeresis Reverse Identity

$$\sin(x)-\sin(y)=2\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)$$

with $x=\frac1{n+1}$ and $y=\frac1n$ reveals for $n\ge 1$

$$\sin\left(\frac1{n+1}\right)-\sin\left(\frac1n\right)=-2\sin\left(\frac{1}{2n(n+1)}\right)\cos\left(\frac{2n+1}{2n(n+1)}\right)<0$$

And we are done!

The derivative is

$$ - (1/n^2) \cos (1/n) $$

As long as $n>1$ the first factor is always positive and so is the second one so the entire thing is negative.

You don't need to prove that $\sin 1/n$ is decreasing: $$\sin\frac1n=\frac1n+a_n$$ where $a_n=O(n^{-3})$, so that $$\sum(-1)^n\sin\frac1n=\sum(-1)^n\frac1n+\sum(-1)^na_n.$$ The former sum is convergent, by Leibniz, and the latter sum is absolutely convergent