This question has already an answer here that I unfortunately didn't understand....
First, I proved pointwise convergence easily after expanding $sin(\frac{1}{nx})$ ( taylor expansion near zero ).
When it came to uniform convergence, I did this:
$|R_n(x)| = |\sum_{k=n+1}^{\infty} \frac{(-1)^{k+1}}{k}\sin\frac{1}{kx}| \leq |\frac{\sin\frac{1}{(n+1)x}}{n+1}| \leq \frac{1}{n+1} \longrightarrow 0$.
Please WHY doesn't it converge uniformly on $(0,\infty)$??
I saw an answer that it DOES converge uniformly on $[A,\infty)$, but what is the problem in our domain?!!
When I saw this question, I felt disappointed as I have an exam tomorrow!
I think the main problem is that our doctor gave us only 2 ways ( normal convergence and remainder term ) to check the uniform convergence despite the fact they're many!! Please help me.
It appears that you are trying to use the Alternating Series Test on $$ \sum_{k=n+1}^\infty\frac{(-1)^{k+1}}{k}\,\sin\left(\frac1{kx}\right)\tag{1} $$ to get that the absolute value of the sum is no greater than the absolute value of the first term, which is no greater than $\frac1{n+1}$.
Unfortunately, the Alternating Series Test requires the terms to be alternating in sign and decreasing in absolute value; that is not the case in $(1)$ when $x$ gets close to $0$.