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?
 A: You want to compute 
$$\lim_{n \rightarrow +\infty} \prod_{k=1}^n \frac{2k-1}{2k}$$
You have that 
$$ \ln \left( \prod_{k=1}^n \frac{2k-1}{2k} \right) = \sum_{k=1}^n \ln \left( 1 - \frac{1}{2k} \right)$$
Moreover, $$\ln \left(1-\frac{1}{2k}\right) \sim - \frac{1}{2k}$$
which is the general term of a divergent series.
Therefore, the series $\sum \ln \left( 1 - \frac{1}{2k} \right)$ diverges to $-\infty$. So your original product converges to $0$. 
A: Let
$$
a_n=\frac12\frac34\cdots\frac{2n-1}{2n}
$$
and
$$
b_n=\frac23\frac45\cdots\frac{2n}{2n+1}
$$
Then, because
$$
\frac k{k+1}\lt\frac{k+1}{k+2}
$$
we have
$$
a_n\lt b_n
$$
and therefore,
$$
a_n^2\lt a_nb_n=\frac1{2n+1}
$$
Thus,
$$
a_n\lt\frac1{\sqrt{2n+1}}
$$
A: With a Wallis' integral: 
It is known (and it can be proved by induction) that
$$\int_0^{\tfrac\pi 2}\sin^{2n}x\,\mathrm dx= \frac\pi2\,\frac{1\cdot 3\,\dots (2n-1)}{2\cdot 4\,\dots (2n)}$$
so $$\frac{1\cdot 3\,\dots (2n-1)}{2\cdot 4\,\dots (2n)}=\frac2\pi\int_0^{\tfrac\pi 2}\sin^{2n}x\,\mathrm dx,$$
and the integral converges to $0$ by the dominated convergence theorem.
A: 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
$$\displaystyle \lim_{n\rightarrow\infty}(1-A_n+(-1)^{n} \displaystyle \prod_{j=1}^{n}\frac{1}{2j})=1-1 +0=0
$$
