Suppose we have the following sequence of functions: $f_n:[0,1]\to \mathbb{R}$
$$ f_n(x)=nx(1-x)^n$$
I need to show this sequence does not converge uniformly to $f(x)=0$
For my definition of uniform convergence I am using:
$f_n,f:I\to \mathbb{R} \\ \qquad f_n \rightrightarrows f \iff \text{if} \enspace \forall \epsilon \enspace \exists n_0 \enspace \text{such that} \enspace \forall n\ge n_0 \enspace \text{then}:$
$$\sup_{x\in I}|f_n(x)-f(x)|<\epsilon$$
What I've done: (I'm not sure if it is correct)
If I fixate n=1, clearly we have $f_1(x)=1-x^2$, in this case, I could say
$$ \sup_{x\in [0,1]}|1-x^2-0|\ge 1$$
Does this show our sequence of functions does not converge uniformly to the 0 function? I guess I'm having trouble taking the negation of the definition of uniform convergence.