I wish to show the uniform convergence of the $f_n$ to $f$ on the interval $[1, \infty)$ where $$f_n(x) = (n/x^3)(\sin(x/n))$$ and $$f(x) = 1/x^2.$$
My idea is as follows: First, I show that $f-f_n$ is always nonnegative.
Second, I show that $(f-f_n)’$ is negative for any $x\geq 1$.
From here, we see that $f-f_n$ is decreasing on $[1,\infty)$. Thus, $f-f_n$ obtains its maximum value in $[1,\infty)$ at 1, namely $ 1- n \sin(1/n)$. Since this value tends to 0 as $n$ approaches infinity and for each $n$ this is the maximum difference between $f_n$ and $f$, the convergence is uniform.
However, I am stuck in showing the derivative is negative for $x\geq 1$. This is equivalent to showing that $$ (3 n \sin(x/n) - x (\cos(x/n) + 2))<0 $$ for $x\geq 1$ and $n \in \mathbb{N}$.
Any help would be appreciated.