= Unbounded multipliers can preserve uniform convergence
{title2=$g(x)=1/x,\quad0<x\leq1/2$}
An unbounded <function> can multiply a <series> while preserving <uniform convergence>. On $E=(0,1/2]$, let $g(x)=1/x$. The <geometric series> gives
$$
\sum_{n=0}^{\infty}x^ng(x)=\frac1{x(1-x)},\qquad
\sup_{x\in E}\left|\frac1{x(1-x)}-\sum_{n=0}^Nx^{n-1}\right|=2^{1-N}\longrightarrow0.
$$
The equality follows because $x^N/(1-x)$ increases up to $x=1/2$. Both the first term and the sum are unbounded, while their difference after sufficiently many terms is uniformly small. <Uniform convergence> controls approximation errors; boundedness of each individual <function> is a separate property.
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