Solution
= Solution
For each fixed $0<x<1$, $x^n\to0$, so the <pointwise limit> is $f(x)=1/(1-x)$. However
$$
|f_n(x)-f(x)|=\frac{x^n}{1-x}\longrightarrow\infty\quad\text{as }x\uparrow1
$$
for every fixed $n$. Hence the error has infinite <supremum> on $(0,1)$ and \b[the convergence is not uniform]. This does not contradict <uniform convergence> on any smaller closed interval $[0,r]$, $r<1$, where the error is at most $r^n/(1-r)$.