 Solution

The finite geometric series gives $F_n(x)=(1-x^{n+1})/(1-x)$, so the pointwise limit is $F(x)=1/(1-x)$. The error is $x^{n+1}/(1-x)$, unbounded as $x\uparrow 1$ for every fixed $n$. Thus convergence is not uniform on $(0,1)$. It is uniform on any smaller interval bounded away from one, since the error is at most $b^{n+1}/(1-b)$ for $0<x\le b<1$.