<s>[INST] Find the limit of $(\frac{1}{2} \frac{3}{4} ...... \frac{2n -1}{2n})$? Find the limit of $(\frac{1}{2} \frac{3}{4} ...... \frac{2n -1}{2n})$?
I have calculated it through calculating the limit of the general term $a_{n} = \frac{2n -1}{2n}$ and it was equal to 1, am I correct?
 [/INST] I. The limit of $a_n$
$$
a_n=\frac{2n-1}{2n}\implies a_n=1-\frac{1}{2n}
$$
$$\displaystyle \lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}(1-\frac{1}{2n})=1
$$

II. The limit of your product (an approximation)
$$\frac{1}{2}\cdot \frac{3}{4}\cdot …\cdot\frac{2(n-1) -1}{2(n-1)} \cdot\frac{2n -1}{2n}=(1-\frac{1}{2})(1-\frac{1}{4})\cdot …\cdot(1-\frac{1}{2(n-1)})(1-\frac{1}{2n})
$$
The right side equals into:
$$
[1-0-\frac{1}{2}](1-\frac{1}{4})\cdot …\cdot(1-\frac{1}{2(n-1)})(1-\frac{1}{2n})
$$
$$
[1-\frac{3}{4}+\frac{1}{2}\cdot\frac{1}{4}](1-\frac{1}{6})\cdot …\cdot(1-\frac{1}{2(n-1)})(1-\frac{1}{2n})=
$$
$$[1-\frac{2}{3}-\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{6}](1-\frac{1}{8})\cdot …\cdot(1-\frac{1}{2(n-1)})(1-\frac{1}{2n})=
$$
$$[1-\frac{35}{48}+\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{6}\cdot\frac{1}{8}](1-\frac{1}{10})\cdot …\cdot(1-\frac{1}{2(n-1)})(1-\frac{1}{2n})=
$$
$$[1-\frac{2887}{3840}-\frac{1}{10!!}](1-\frac{1}{12})\cdot …\cdot(1-\frac{1}{2(n-1)})(1-\frac{1}{2n})=
$$
$$...
$$
$$=1-A_n+(-1)^{n} \displaystyle \prod_{j=1}^{n}\frac{1}{2j}
$$
What I can say about $A_n$ at least is
$$A_n\rightarrow 1+\frac{(-1)^{n}}{2n!}
$$
Hence
$$\displaystyle \lim_{n\rightarrow\infty}(1-A_n+(-1)^{n} \displaystyle \prod_{j=1}^{n}\frac{1}{2j})=1-1 +0=0
$$
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