(b) Show $(f_n)$ does not converge uniformly.
Proof: (a) First, note that $f(x)=\sum kx^k$ converges for all $|x| < 1$ by the ratio test. Since each compact subset of $(-1,1)$ is contained in some interval $[-x,x]⊆(-1,1)$, it will suffice to show $f_n$ converges uniformly on $[-x,x]$ for all $x∈(0,1)$. Therefore let $|y| ≤ |x|$ and observe $$|f(y)-f_n(y)| = |\sum^∞_{k=1}ky^k-\sum^n_{k=1}ky^k| = |\sum^∞_{k=n+1} ky^k| ≤$$$$\sum^∞_{k=n+1} k|x|^k →0$$ as $n→∞$.
(b) Suppose some $n$ such that $|f(x)-f_n(x)| < 1/2$ for all $x∈(-1,1)$. Simply choose $x∈(0,1)$ so that $(n+1)x^{n+1} ≥ 1/2$, and observe $$|f(x)-f_n(x)| = |\sum_{k=n+1}^∞ kx^k| ≥ 1/2~~\square$$
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