Solution
= Solution
{c}
The finite <geometric series> identity gives, for each $|x|<1$,
$$
f_n(x)=\frac{x(1-x^n)}{1-x}\longrightarrow f(x)=\frac{x}{1-x},\qquad f(x)-f_n(x)=\frac{x^{n+1}}{1-x}.
$$
For every fixed $n$, the <absolute value> of this error tends to infinity as $x\to1^-$. Thus
$$
\boxed{\sup_{-1<x<1}|f(x)-f_n(x)|=\infty,}
$$
and there is \b[no <uniform convergence>] on $(-1,1)$.