(b) Show (fn) does not converge uniformly.
Proof: (a) First, note that f(x)=∑kxk 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 fn converges uniformly on [−x,x] for all x∈(0,1). Therefore let |y|≤|x| and observe |f(y)−fn(y)|=|∞∑k=1kyk−n∑k=1kyk|=|∞∑k=n+1kyk|≤∞∑k=n+1k|x|k→0 as n→∞.
(b) Suppose some n such that |f(x)−fn(x)|<1/2 for all x∈(−1,1). Simply choose x∈(0,1) so that (n+1)xn+1≥1/2, and observe |f(x)−fn(x)|=|∞∑k=n+1kxk|≥1/2 ◻
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