Can someone verify my work here?
So I'm given the sequence of functions $f_n(x)=\frac{2(-1)^{n+1}sin(n \pi x)}{n \pi}$ and I need to determine whether it is uniformly convergent on $[0,1]$.
i) 2 and $\pi$ are constant so I proceed with the following:
$0\le(-1)^{n+1}sin(n\pi x)\le 1 $
$\frac{0}{n}\le\frac{(-1)^{n+1}sin(n\pi x)}{n}\le \frac{1}{n} $
$\lim_{n\rightarrow \infty}\frac{0}{n}\le\lim_{n\rightarrow \infty}\frac{(-1)^{n+1}sin(n\pi x)}{n}\le \lim_{n\rightarrow \infty}\frac{1}{n} $
$0\le \lim_{n\rightarrow \infty} f_n(x)\le 0 $
$\therefore \lim_{n\rightarrow \infty} f_n(x) = 0$
ii) Now , let $\epsilon >0$, then $\big|f_n(x)-0\big|=\big|\frac{2(-1)^{n+1}sin(n \pi x)}{n \pi}\big| \le \frac{2\cdot 1}{\frac{n}{2}}=\frac{4}{n}<\epsilon$.
Thus, $\forall \;\epsilon>0,\;\exists $ a corresponding number $N=\frac{4}{\epsilon}$ such that $|f_n(x)-f(x)|<\epsilon $ for all $x \in [0,1]$ and all $n>N$.
Hence $f_n(x)$ converges uniformly to $0$ on $[0,1]$.
Your "$0 \leq (-1)^{n+1} \sin(n \pi x) \leq 1$" line is wrong. For example $n=3$ and $x=1/2$. In general for $n=4k - 1$, $k=1,2, \dots$ and $x=1/2$. So you need to use absolute value instead.
$\displaystyle \left| f_n(x) \right| = \left| \frac{2 (-1)^{n+1} \sin (n \pi x)}{n \pi} \right| \leq \frac{2}{n \pi} \left| (-1)^{n+1} \sin (n \pi x) \right| \leq \frac{2}{n \pi} \to 0$ as $n \to \infty$.
So given $\epsilon>0$, choose $N \in \mathbb{N}$ such that $N>\dfrac{2}{\pi \epsilon}$ then for $n>N$ you have $\displaystyle \left| f_n(x) - 0 \right| \leq \frac{2}{n \pi} < \frac{2}{N \pi} < \epsilon$
for all $x \in [0,1]$. Hence $f_n$ converges to $0$ uniformly on $[0,1]$. (In fact on $\mathbb{R}$.)