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Question Number 195393 by Erico last updated on 01/Aug/23

prove that

\lim_{x \to 0} \sqrt[x]{\frac{\underset{k = 1}{\sum^{n}} {(1 - \frac{1}{2 k})}^{x}}{n}} = \frac{1}{4} \sqrt[n]{C_{2 n}^{n}}

Answered by witcher3 last updated on 01/Aug/23

f (t) = t^{x} = e^{xln (t)} = 1 + xln (t) + o (x), x \to 0

\underset{k = 1}{\sum^{n}} {(1 - \frac{1}{2 k})}^{x} = Σ (1 + xln (1 - \frac{1}{2 k}) + o (x))

= n + xln (\underset{k = 1}{\prod^{n}} (\frac{2 k - 1}{2 k})) + o (x)

= n + xln (\underset{k = 1}{\prod^{n}} \frac{(2 k) (2 k - 1)}{{4 k}^{2}}) + o (x)

= n + xln (\frac{(2 n)!}{4^{n} {(n!)}^{2}}) + o (x) = n + xln (\frac{1}{4^{n}} C_{2 n}^{n}) + o (x)

\lim_{x \to 0} \sqrt[x]{\frac{1}{n} \underset{k = 1}{\sum^{n}} {(1 - \frac{1}{2 k})}^{x}} = {\underset{x \to 0}{\lim e}}^{\frac{1}{x} \ln (\frac{1}{n} (n + xln (\frac{1}{4^{n}} C_{2 n}^{n}) + o (x))}

= {\underset{x \to 0}{\lim e}}^{\frac{\ln (1 + \frac{x}{n} \ln (\frac{1}{4^{n}} C_{2 n}^{n}) + o (1))}{x}} = e^{\lim_{x \to 0} \frac{\ln (1 + \frac{x}{n} \ln (\frac{C_{2 n}^{n}}{4^{n}}) + o (1))}{x}}

≪ \exp is continus ≫

= e^{\lim_{x \to 0} \frac{\frac{x}{n} \ln (\frac{C_{2 n}^{n}}{4^{n}}) + o (1)}{x}} = e^{\ln (\sqrt[n]{\frac{C_{2 n}^{n}}{4^{n}}})} = \sqrt[n]{\frac{C_{2 n}^{n}}{4^{n}}}

\frac{1}{4} \sqrt[n]{C_{2 n}^{n}}

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