$ \lim_{n\to+\infty} \frac{1\times 3\times \ldots \times (2n+1)}{2\times 4\times \ldots\times 2n}\times\frac{1}{\sqrt{n}}$

Question

Sabiendo que : $$I_n=\int_0^{\frac{\pi}{2}}\cos^n(t) \, dt$$

$$I_{2n}=\frac{1\times 3\times \ldots \times (2n-1)}{2\times 4\times \ldots\times 2n}\times\dfrac{\pi}{2}\quad \forall n\geq 1$$ $$I_{n}\sim \sqrt{\dfrac{\pi}{2n}}$$

Calcular: $$ \lim_{n\to+\infty} \frac{1\times 3\times \ldots \times (2n+1)}{2\times 4\times \ldots\times 2n}\times\dfrac{1}{\sqrt{n}}$$

De hecho,

$$I_{2n}=\frac{1\times 3\times \ldots \times (2n-1)}{2\times 4\times \ldots\times 2n}\times\dfrac{\pi}{2}\quad \forall n\geq 1 \\ \frac{1\times 3\times \ldots \times (2n-1)}{2\times 4\times \ldots\times 2n}=\dfrac{2}{\pi}\times I_{2n}$$

entonces

$$ \begin{align*} \frac{1\times 3\times \cdots \times (2n+1)}{2\times 4\times \cdots\times 2n}\times\dfrac{1}{\sqrt{n}}&=\dfrac{2}{\pi}\times I_{2n}\times (2n+1)\times\dfrac{1}{\sqrt{n}}\\ &=\dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{\sqrt{n}} \times \sqrt{ \dfrac{2n\times I^{2}_{2n}}{2n} } \\ &=\dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{ \sqrt{2}\times n} \times \sqrt{ 2n\times I^{2}_{2n} } \\ \end{align*} $$ o $$(2n)I^{2}_{2n}\sim \dfrac{\pi}{2}$$

entonces

\begin{align*} \frac{1\times 3\times \cdots \times (2n+1)}{2\times 4\times \cdots\times 2n}\times\dfrac{1}{\sqrt{n}}&=\dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{ \sqrt{2}\times n} \times \sqrt{ 2n\times I^{2}_{2n} } \\ &\sim\dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{ \sqrt{2}\times n} \times \sqrt{ \dfrac{\pi}{2} } \\ \end{align*} $$ Estoy atascado aquí

creo que puedo seguir adelante

\begin{align*} \frac{1\times 3\times \cdots \times (2n+1)}{2\times 4\times \cdots\times 2n}\times\dfrac{1}{\sqrt{n}}&\sim \dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{ \sqrt{2}\times n} \times \sqrt{ \dfrac{\pi}{2} } \\ &\sim \dfrac{2}{\pi}\times\dfrac{2n}{ \sqrt{2}\times n} \times \sqrt{ \dfrac{\pi}{2} } \\ &\sim \frac{ 2\sqrt{\pi} }{\pi}=\dfrac{2}{\sqrt{\pi}} \end{align*} estoy en lo cierto ? si que lo es hay alguna otra manera

Answer 1

Así que sabemos que:$$ I_{2n}=\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta = \frac{\pi}{2}\cdot\frac{(2n-1)!!}{(2n)!!}=\frac{\pi}{2}\cdot\frac{1}{4^n}\binom{2n}{n}.\tag{1}$ $ Si, de la misma manera, se prueba:$$ I_{2n+1}=\int_{0}^{\pi/2}\cos^{2n+1}(\theta)\,d\theta = \frac{(2n)!!}{(2n+1)!!}\tag{2} $ $ entonces tenemos:

$$\lim_{n\to +\infty} n\,I_{2n}\, I_{2n+1} = \lim_{n\to +\infty}\frac{\pi n}{4n+2}=\frac{\pi}{4}.\tag{3}$ $ Desde$\{I_n\}_{n\in\mathbb{N}}$ es una secuencia decreciente, disponemos$\frac{I_{2n+1}}{I_{2n}}\leq 1$, así como:$$ \frac{I_{2n+1}}{I_{2n}}\geq \frac{I_{2n+2}}{I_{2n}} = \frac{2n+1}{2n+2}\geq 1-\frac{1}{2n}\tag{4}$ $, por lo tanto, apretando:$$ \lim_{n\to +\infty} n I_{2n}^2 = \frac{\pi}{4}, \tag{5}$ $ a continuación:

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Answer 2

Se nos da

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y

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De$$I_{2n}=\frac{\pi}{2}\frac{(2n-1)!!}{(2n)!!} \tag 1$ es trivial ver que

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Luego, utilizando$$I_n\sim \sqrt{\frac{\pi}{2n}} \tag 2$ y$(2)$, nos encontramos con que

$$ \begin{align} \frac{(2n+1)!!}{\sqrt{n}(2n)!!}&=\frac{(2n+1)(2n-1)!!}{\sqrt{n}(2n)!!}\\\\ &=\frac{2n+1}{\sqrt{n}}\left(\frac{(2n-1)!!}{(2n)!!}\right)\\\\ &=\frac{2n+1}{\sqrt{n}}\left(\frac{2}{\pi}I_{2n}\right)\\\\ &\sim \frac{2n+1}{\sqrt{n}}\left(\frac{2}{\pi}\frac12\sqrt{\frac{\pi}{n}}\right)\\\\ &=\frac{2n+1}{n\sqrt{\pi}}\\\\ &\to \frac{2}{\sqrt{\pi}}\,\,\text{as}\,\,n\to \infty \end {Align} $$

Por lo tanto, tenemos

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