= Solution
{c}
Choose the unbounded <function> $\boxed{g(x)=1/x}$ on $E=(0,1/2]$. It is unbounded because $g(1/k)=k$ for all <integers> $k\geq2$. For each $x\in E$, the <geometric series> gives the sum and its remainder after degree $N$:
$$
\sum_{n=0}^{\infty}x^ng(x)=\frac1{x(1-x)},
\qquad
\frac1{x(1-x)}-\sum_{n=0}^Nx^{n-1}=\frac{x^N}{1-x}.
$$
Since $0<x\leq1/2$,
$$
\boxed{\sup_{x\in E}\left|\frac1{x(1-x)}-\sum_{n=0}^Nx^{n-1}\right|\leq2^{1-N}\longrightarrow0.}
$$
This proves <uniform convergence>. Although the first term $1/x$ and the sum are unbounded, the late tails have a bound tending to zero independently of $x$. <Uniform convergence> concerns those tails and does not require the individual <functions> to be bounded.
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