Compute $\lim\limits_{n \to \infty} {\frac{1 \cdot 3 \cdot 5 \cdots(2n - 1)}{2 \cdot 4 \cdot 6 \cdots (2n)}}$ EDIT: @Holo has kindly pointed out that my concept of ln rules used in this question is wrong. However, the intuition behind using the tangent of a curve to find the sum to infinity of a series still stands. Therefore, I won't be editing the post.
I tried expanding out the equation from the question and got $${a_{n}} = \frac{1}{2} .\frac{3}{4} . \frac{5}{6} ...\frac{2n -1}{2n}$$
I then tried taking the ln of the equation which works out to $$\ln(1 - \frac{1}{2}) + \ln(1 - \frac{1}{4}) + ...$$
Here is my question. I used the rules of ln functions and took $ln(\frac{1}{1/2}) + ln(\frac{1}{\frac{1}{4}})$ + ...
Since ln(1) is 0, shouldn't the limit work out to 0 as n tends to infinity? Then $$ln(L) = 0, where L = limit$$
$$L = e^0$$
$$L = 1$$ 
However, the limit is actually 0, and my tutor used a method which I couldn't understand as he tried approximating the function $y = ln(x)$ to $y = x - 1$, as he says that the linear equation is actually a tangent to the ln curve. Could someone please explain this intuition behind it? He differentiated the equation, and since the curve cuts the x-axis at x = 1, he got the linear curve $y = x - 1$.
He did mention that the linear equation is just an approximation, but said such an approximation would be more than sufficient.
 A: A completely different approach is to write
$${a_{n}} = \frac{1}{2} .\frac{3}{4} . \frac{5}{6} ...\frac{2n -1}{2n}=\frac {(2n)!}{(2^nn!)^2}$$
because you can divide a $2$ out of each term on the bottom and get $n!$ and you can multiply top and bottom by the bottom.  Now feed it to Stirling
$$a_n\approx\frac{(2n)^{2n}e^{2n}}{e^{2n}2^{2n}n^{2n}\sqrt{\pi n}}=\frac 1{\sqrt{\pi n}}\to 0$$
A: As you say,
$$\ln a_n=\sum_{k=1}^n\ln\left(1-\frac1{2k}\right).$$
But as $x\to0$,
$$\ln(1-x)=-x+O(x^2).$$
Therefore
$$\ln a_n=-\sum_{k=1}^n\frac1{2k}+O\left(\sum_{k=1}^n\frac1{k^2}\right).$$
As the series $\sum_1^\infty1/k$ diverges and $\sum_1^\infty1/k^2$ converges,
then $\ln a_n\to-\infty$, and so $a_n\to0$.
A: We can prove that 
$${a_{n}} = \frac{1}{2} .\frac{3}{4} . \frac{5}{6} ...\frac{2n -1}{2n}<\frac{1}{\sqrt{2n+1}}.$$
Let $b_n=\frac{2}{3} .\frac{4}{5} . \frac{6}{7} ...\frac{2n}{2n+1}$, then it is easy to see that $a_n<b_n$. So $$a_n^2<a_nb_n=\frac{1}{2n+1}.$$
You limit is $$\lim_{n\to \infty}\frac{1}{2} .\frac{3}{4} . \frac{5}{6} ...\frac{2n -1}{2n}=0.$$
