Convergence point of $\frac{\prod_{k=1}^n (2k-1)} {\prod_{k=1}^n 2k}$ Let $a_n = \frac{\prod_{k=1}^n (2k-1)} {\prod_{k=1}^n 2k}$ is a sequence of numbers . Is $a_n$ monotone and bounded ? If it's a convergent sequence find convergence point .
My try : It's obvious that $a_n$ is decreasing and bounded. So we can deduce it's a convergent sequence but I'm not able to find $\lim_{n \to \infty} a_n$ .
 A: HINT:
$a_n < b_n \colon = \prod_{k=1}^n \frac{2k}{2k+1}$ so 
$a_n < \sqrt{a_n b_n} = \sqrt{\frac{1}{2n+1}}$
A: If you know Wallis's Product, then you can use the approximation
$$\frac{\pi}{2}\approx \prod_{k=1}^n\,\left(\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}\right)$$
to show that
$$\frac{\pi}{2} \approx \frac{1}{2n+1}\,\prod_{k=1}^n\,\left(\frac{2k}{2k-1}\right)^2\,.$$
That is,
$$\prod_{k=1}^n\,\frac{2k-1}{2k}\approx\frac{1}{\sqrt{\pi\left(n+\frac{1}{2}\right)}}\,.$$
A: $$\prod_{k=1}^n \frac{2k-1}{2k}=\left(\prod_{k=1}^n \frac{2k}{2k-1}\right)^{-1}=\left(\prod_{k=1}^n \left(1+\frac{1}{2k-1}\right)\right)^{-1}$$
However, we know that 
$$1+\sum_{k=1}^n\frac1{2k-1} \leq \prod_{k=1}^n \left(1+\frac{1}{2k-1}\right)$$
Since $\lim_{n \to \infty}\sum_{k=1}^n\frac1{2k-1}=\infty$,
We have  $\lim_{n \to \infty}\prod_{k=1}^n \left(1+\frac{1}{2k-1}\right)=\infty$
Hence 
$$\prod_{k=1}^n \frac{2k-1}{2k}=\left(\prod_{k=1}^n \frac{2k}{2k-1}\right)^{-1}=\left(\prod_{k=1}^n \left(1+\frac{1}{2k-1}\right)\right)^{-1}=0$$
A: $a_n=\prod_{k=1}^n(1-b_k)$ where $b_k=\frac {1}{2k}.$ Apply part [1] of the following :
Theorem. If $0\leq b_k<1$ for all $k$ then  $\prod_{k=1}^{\infty}(1-b_k)>0 \iff \sum_{k=1}^\infty b_k<\infty$.
Proof:
[1].  $\ln (1-x)\leq (-x)$ for $0\leq x<1$ so $\ln \prod_{k=1}^n(1-b_k)\leq \sum_{k=1}^n (-b_k).$ 
So if $\sum_{k=1}^{\infty}b_k=\infty$ then $\ln \prod_{k=1}^n(1-b_k)\to -\infty$ as $n\to \infty.$
[2]. If $\sum_{k=1}^{\infty}b_k<\infty$:  Let $a_k=\frac {b_k}{1-b_k}.$ Then $a_k\geq 0$ and $(1+a_k)^{-1}=1-b_k.$   Now $b_k\leq 1/2$ for all but finitely many $k$ because $\sum_{k=1}^{\infty}b_k<\infty.$ So for all but finitely many $n$ we have $$a_n=\frac {b_k}{1-b_k}\leq\frac {b_k}{1-\frac {1}{2}}=2b_k.$$  We have $\ln (\;(1+x)^{-1})\geq (-x)$ for $x\geq 0$ so $$\ln \prod_{k=1}^n(1-b_k)=\ln \prod_{k=1}^n(1+a_k)^{-1}\geq  \sum_{k=1}^n(-a_k)\geq$$ $$\geq  \sum_{k\in \mathbb N: b_k\leq 1/2}(-2b_k)+\sum_{k\in \mathbb N:b_k> 1/2}(-a_k).$$ The first summation in the line above is bounded below by $-2\sum_{k=1}^{\infty}b_k$ and the second summation  has finitely many terms, so there is a common finite lower bound for $\ln \prod_{k=1}^n(1-b_k)$ over all $n.$
Reamrk: There is a similar theorem: If $c_k\geq 0$ for all $k$ then $\prod_{k=1}^{\infty}(1+c_k)<\infty \iff \sum_{k=1}^{\infty}c_k<\infty$. Both these theorems can be proven without logarithmic inequalities but it is much easier to use them.
