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Prove that the series $\sum_{n=1}^{\infty}\left\Vert x\right\Vert ^{n} $, $x\in\mathbb{R}^{n} $, does not converge uniformly on the unit ball $\left\{ x\in\mathbb{R}^{n}\mid\left\Vert x\right\Vert <1\right\} $.

I am not sure how to show this. What I got to is that the given series is a geomtric series and hence $$f\left(x\right)=\sum_{n=1}^{\infty}\left\Vert x\right\Vert ^{n}=\frac{\left\Vert x\right\Vert }{1-\left\Vert x\right\Vert }$$ on the unit ball, which is continuous (on the unit ball). But this doesn't tell us anything.

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    $\begingroup$ Consider, for fixed $m$, $\sum_{n=m}^\infty\Vert x\Vert^n={\Vert x\Vert^m\over 1-\Vert x\Vert}$ and show this can't be uniformly small over the unit ball. $\endgroup$ – David Mitra Jul 27 '12 at 0:02
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Suppose this converges uniformly to $\frac{\|x\|}{1-\|x\|}$. Then for any $\epsilon>0$, we have some $N$ such that $$\frac{\|x\|}{1-\|x\|}-\sum\limits_{n=1}^N \|x\|^n<\epsilon, \forall x\in B_1(0)$$ but this cannot be the case, as $$\lim\limits_{\|x\|\to 1}\frac{\|x\|}{1-\|x\|}-\sum\limits_{n=1}^N \|x\|^n=\infty>\epsilon$$ since the sum is bounded by $N$ yet the fraction diverges to $\infty$. Thus the series does not converge uniformly on the unit ball $B_1(0)$.

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Each partial sum is bounded on the unit ball. Uniform convergence preserves boundedness. The sum is unbounded.

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