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Prove that for every $n\in\mathbb{N}$ there exists a polynomial $P$ such that $\deg P=n$ and $$ \sup_{x\in\left[0,\frac{1}{2}\right]}\left| P(x)-\frac{1}{1-x}\right|\ <\ \frac{4}{(\sqrt{2}+1)^{2n+2}} $$

I tried with the incomplete $\frac{1}{1-x}$, i.e. $1+x+\dots+x^n$, but it doesn't work for $x=\frac{1}{2}$ and close values.

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    HIint (I haven't tried). Consider the Taylor series about $x=1/4$. – 2017-02-24
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    correct me if I'm wrong: the $k$-th derivative of $\frac{1}{1-x}$ seems to be $\frac{k!}{1-x)^{k+1}}$, so for such $P$ we would have $LHS=\frac{4}{3}\sum_{k=n+1}^{\infty}\frac{1}{3^k}=\frac{2}{3}\frac{1}{3^n}$ what is bigger then needed – 2017-02-24
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    Tight bounds for absolute errors of polynomial approximations usually lead to answers involving Tchebyschev polynomials. – 2017-03-05

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