How find x if $\frac{1}{x}+\frac{2}{x^3}+\frac{6}{x^5}+\cdots+\frac{\binom{2n}{n}}{x^{2n+1}}+\cdots=1$

Asked Dec 30 '14 at 12:36

Active Dec 30 '14 at 12:52

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Let $x>2$ and $\frac{1}{x}+\frac{2}{x^3}+\frac{6}{x^5}+\cdots+\frac{\binom{2n}{n}}{x^{2n+1}}+\cdots=1$. How find $x$?

asked Dec 30 '14 at 12:36

piteer

2

Does the fact $$ \sum_{n=0}^{\infty} \binom{2n}{n}z^{n} = (1-4z)^{-1/2}, \quad |z| < 1/2$$ help you? – Sangchul Lee Dec 30 '14 at 12:40

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