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$\mathbf{Question:}$ Show that $\frac{(1)(3)(5)\dots(2n-1)}{(2)(4)(6)\dots(2n)}$ is convergent where $n\in \mathbb{N}$

$\mathbf{My\ attempt:}$ Let $a_n = \frac{2n-1}{2n}$ and let $f(n) = a_n$

$$ f(n)=\frac{2n-1}{2n} = 1-\frac{1}{2n} $$

$$ f'(n) = \frac{1}{2n^2} $$

As $f'(n)>0$, it is a strictly increasing sequence

And $\frac{2n-1}{2n} >0$, therefore it is bounded below

But according to the Monotone convergence theorem, this sequence is divergent instead of convergent?

Any help is appreciated.

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    $\begingroup$ The sequence you are referring to $a_n$ is not the same as the sequence in your question. So saying anything about $a_n$ is not of much help. $\endgroup$
    – Anurag A
    Commented Jun 26, 2020 at 23:02
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    $\begingroup$ Your given sequence (based on your question) is $$a_n=\frac{1\cdot 3 \cdot 5 \dotsb (2n-1)}{2 \cdot 4 \cdot \cdot 6\dotsb(2n)}=\frac{1\cdot \color{red}{2} \cdot 3 \cdot \color{red}{4} \cdot 5 \dotsb (2n-1) \cdot \color{red}{(2n)}}{2^2 \cdot 4^2 \cdot 6^2\dotsb(2n)^2}=\frac{(2n)!}{2^{2n}(n!)^2}$$ $\endgroup$
    – Anurag A
    Commented Jun 26, 2020 at 23:08
  • $\begingroup$ There's a wrinkle in terminology that makes the wording of this question a little unfortunate. The limit being asked about that of an infinite product, $\ \displaystyle\prod_{i=1}^\infty\left(1-\frac{1}{2i}\right)\ $, whose limit happens to be zero. When the limit of an infinite product with no zero terms is nevertheless zero, it is said to "diverge to zero", even though the sequence $\ a_n\ $ defined by $\ \displaystyle a_n=\prod_{i=1}^n\left(1-\frac{1}{2i}\right)\ $ is actually convergent. $\endgroup$ Commented Jun 26, 2020 at 23:36

2 Answers 2

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We define:

$$a_n=\prod_{k=1}^n\bigg(\frac{2k-1}{2k}\bigg)$$ Then, $$a_{n+1}=\prod_{k=1}^{n+1}\bigg(\frac{2k-1}{2k}\bigg)=a_n\cdot\frac{2n+1}{2n+2}$$

$$\implies \frac{a_{n+1}}{a_n}=\frac{2n+1}{2n+2}<1$$

$$\implies a_{n+1}<a_{n}$$ $$(a_n)_{n\in \Bbb N} \text{ converges}$$

For boundedness, observe that $a_1=\frac12$ and each element of our product is non-negative, thusly $0<a_j\leq \frac12$ for each $a_j$

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    $\begingroup$ You still need to show boundedness as well to claim convergence. Also the last line should be $a_{n+1}<a_n$. $\endgroup$
    – Anurag A
    Commented Jun 26, 2020 at 23:12
  • $\begingroup$ Good spot, thank you. $\endgroup$ Commented Jun 26, 2020 at 23:17
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Since $\dfrac{a_{n+1}}{a_n} =\dfrac{2n+1}{2n+2} =1-\dfrac{1}{2n+2} $,

$\begin{array}\\ f(n, m) &=\dfrac{a_{m+1}}{a_n}\\ &=\prod_{k=n}^m \dfrac{a_{k+1}}{a_k}\\ &=\prod_{k=n}^m (1-\dfrac{1}{2k+2})\\ g(n, m) &=\ln(f(n, m))\\ &=\sum_{k=n}^m \ln(1-\dfrac{1}{2k+2})\\ &\lt\sum_{k=n}^m -\dfrac{1}{2k+2} \qquad\text{since }\ln(1-x) < -x\\ &\to -\infty \qquad\text{as } m \to \infty\\ \text{so}\\ f(n, m) &\to 0 \qquad\text{as } m \to \infty\\ \end{array} $

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