 Solution

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 $k\ge 2$. For each $x\in E$, the geometric series gives the sum and its remainder after degree $N$:Since $0<x\le 1/2$,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.