Prove that $f_n(x) = 1 - \cos(\frac{x}{n})$ is uniformly convergent 
Prove that $f_n(x) = 1 - \cos(\frac{x}{n})$ is uniformly convergent in $[0,2\pi]$.

So I said this, and my professor said it is wrong:

$|1-\cos(\frac{x}{n})| \leq 2, $ then we take some $\epsilon >2 $, and we have for every $n \in \mathbb{N}$ and $\forall x \in [0,2\pi]$ that  $|1-\cos(\frac{x}{n})| < \epsilon$ and it is uniformly convergent.

What is wrong about what I did? 
 A: As I said in my comment, your proof doesn't cover the $\epsilon \in (0,2)$ case, you need to prove for all $\epsilon$ which means also for these.
Hint For $n >4$ you can prove that
$$0<  1 - \cos(\frac{x}{n}) \leq  1 - \cos(\frac{2 \pi}{n}) \forall x \in [0,2 \pi] \,.$$
$n>4$ is needed to make sure we are in the first quadrant. This follows immediately from the monotony of $\cos(x)$ in the first quadrant.
[$n>2$ would actually suffice, but who cares :) ].
Now, use the the definition of $\lim_n \cos(\frac{2 \pi}{n})=1$ and the fact that there is no $x$ in this limit.

Edit Alternately, by the double angle formulas and $\sin(y) <y$:
$$0 \leq 1 - \cos(\frac{x}{n}) = 2 \sin^2(\frac{x}{2n}) \leq 2 (\frac{x}{2n})^2  \leq 2 (\frac{2 \pi}{2n})^2$$
A: Note that $(f_n)$ is pointwise convergent to the zero function in $[0,2\pi]$ moreover we have
$$\forall n\geq 4\quad |f_n(x)|\leq 1-\cos\left(\frac{2\pi}{n}\right),\quad\forall x\in[0,2\pi]$$
so
$$||f_n||_\infty\to0$$
and then we have the uniform convergence of  $(f_n)$ to the zero function.
A: Proposition:  Let $f_n(x) = 1 - \cos(\frac{x}{n})$. Then  $f_n(x) \to 0$ uniformly in $x$ on the interval $[0, 2 \pi]$ as $n \to \infty$.
Proof:  Pick any real $\epsilon > 0$.  Then there exists a real $\delta > 0$  such that $|1 - \cos y| < \epsilon$ if $|y| < \delta$; this assertion follows immediately from the continuity of $\cos y$ and the fact that $\cos 0 = 1$.  Now choose $N > \frac{2 \pi}{\delta}$ and integer $n > N$.  Then for $x \in [0, 2 \pi]$, we have $\frac{x}{n} < \frac{x}{N} \le \frac{2 \pi}{N} < \delta$, whence $f_n(x) < \epsilon$.  Note this holds for any $\epsilon$ and $N$ for all $ x \in [0, 2 \pi]$.  Thus is the uniform convergence of $f_n(x)$ to $0$. established.  QED.
