Another uniform convergence question

Kate2010
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Homework Statement



h:[0,1] -> R is continuous
Prove that t(x) = \sum^{infinity}_{n=0} xnh(xn) is uniformly convergent on [0,s] where 0<s<1

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The Attempt at a Solution



I have the definition of h being continuous but after this I am pretty clueless about how to tackle this problem. I could use the Weierstrass M-test. I know the series xn converges uniformly on this interval as xn < sn but I don't know how to use the fact that h is continuous to find a sequence of real numbers that xnh(xn) is always less than.
 
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Use the extreme value theorem to bound h(x^n), and then find small bounds for x^n h(x^n).
 
h is continuous (it is a continuous function of a polynomial which is continuous) on a compact set therefore h is bounded for x in the interval [0,s]. the sequence of h's for all n is uniformly bounded by a single M. as you said the s^n is a geometric series so it converges. hence uniform convergence.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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