# Convergence in binomial series

Let $r>0$, $\varepsilon>0$ and $\alpha>0$. Assume that $0<\varepsilon<x<r$. I want a power series in $x$ for $x^{\alpha}$. Here is my attempt.
We may assume that $r<1$. $$x^{\alpha}=(1+(x-1))^{\alpha}=\sum_{k=0}^{\infty}\binom{\alpha}{k}(x-1)^k$$ $$=\sum_{k=0}^{\infty}\binom{\alpha}{k}\sum_{j=0}^k \binom{k}{j}x^j(-1)^{k-j}.$$ Now I want to obtain one sum, say, $$=\sum_{n=0}^{\infty} c_n x^n.$$ How could I achieve this and this rearranged series will be uniformly convergent in $[\varepsilon,r]$? (A "little bit" smaller interval is also satisfactory.)

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 Why don't you just use Taylor formula? – Norbert Jul 4 '12 at 15:26 How, precisely, is $\binom\alpha k$ defined for non-integer $\alpha$? The only reasonable way I can think of is with the Gamma function. – Cameron Buie Jul 4 '12 at 15:29 @CameronBuie: $\binom\alpha k=\alpha(\alpha-1)\cdots(\alpha-k+1)/k!$. – Harald Hanche-Olsen Jul 4 '12 at 15:32 Interesting, Harald. Does that work out for a general binomial expansion? – Cameron Buie Jul 4 '12 at 15:39 @Norbert the last power series is about $0$. Calculating the derivative of $x^{\alpha}$ at $0$ you would obtain zero in the denominator. – vesszabo Jul 4 '12 at 16:25

Such a series would converge for $|x|<r$ and that is possible only for integer nonnegative values of $\alpha$.
 right, the power series is convergent in a symmetric interval about $0$, so it will convergent for $|x|0$ is "small", this is the reason of the condition \$0<\varepsilon