Show that the sequence is convergent $\frac{(1)(3)(5)\dots(2n-1)}{(2)(4)(6)\dots(2n)}$

Question

$\mathbf{Question:}$ 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.

Answer 1

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 $a_j$

Answer 2

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} $