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    • Thread Starter

    I'm really struggling with this question:

    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

    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.
    • Study Helper

    Study Helper
    h is continuous on a closed interval implies.....
    • Thread Starter

    Oh yeah, that it is bounded, so h(xn) < M say.

    So xnh(xn) < Msn.
    • Thread Starter

    I'm really struggling with another question now:

    Suppose \sum^{infinity}_{k=0} p(p-1)...(p-k+1)(-1)k/k(k-1)...1 is convergent.
    Show that \sum^{infinity}_{k=0} p(p-1)...(p-k+1)(x)k/k(k-1)...1 is uniformly convergent on [-1,0]

    I have shown that p(p-1)...(p-k+1)(-1)k/k(k-1)...1 < 0 for k=1,2,3,...
    \sum^{infinity}_{k=0} p(p-1)...(p-k+1)(-1)k/k(k-1)...1 = L (< 0) as it converges to a limit.
    |(-1)krk|\leq rk for r<1 and -1<x\leq0
    However, I do not know how to tackle the case when x=-1.
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