If I have understood it correctly; First, let $Q_n = \dfrac{x^{n+1}}{1-x}$.
Then, $$\lim_{n \rightarrow \infty} || Q_n -0|| = \lim_{n \rightarrow \infty} \sup_{x \in (-1,1)}|Q_n|=\lim_{n\rightarrow \infty} \sup_{x \in (-1,1)} \dfrac{|x^{n+1}|}{|1-x|} \qquad \qquad (\ast)$$
For the latter RHS: no matter the value of $n \geq 1$ (just particularly as I am working with sequences), the supremum will always be $\infty$ as,
$$\lim_{x \rightarrow 1^-} \dfrac{x^{n+1}}{1-x} = \dfrac{\text{"1"}}{\text{"0"}} \rightarrow +\infty$$
as no matter what $n$, it will always be infinity. Even less, it is not equal to $0$ in $(\ast)$ as the uniform convergence requires.
Edit: Is this sufficient?