The Formula to this Sequence Series What is the formula for following sequence?
$$\frac12 + \frac12 \cdot \frac34 + \frac12 \cdot \frac34 \cdot \frac56 + ... + \frac12 \cdot \frac34 \cdot \frac56 ... \frac{2n - 1}{2n}$$
This is a question from my Calculus class about sequences.
 A: $$ 
\begin{align} 
&\, 1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1) = (2n-1)!! \,,\quad\, 2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n) = (2n)!! \\[2mm] 
&\, S_{\small n} = \frac{1}{2}+\frac{1\,\cdot\,3}{2\,\cdot\,4}+\frac{1\,\cdot\,3\,\cdot\,5}{2\,\cdot\,4\,\cdot\,6}+\,\cdots\,+\frac{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)}{2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n)} \\[6mm] 
&\, \color{red}{1+S_{\small n}} = 1+\frac{1}{2}+\frac{1\,\cdot\,3}{2\,\cdot\,4}+\frac{1\,\cdot\,3\,\cdot\,5}{2\,\cdot\,4\,\cdot\,6}+\,\cdots\,+\frac{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)}{2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n)} \\[3mm] 
&\, \qquad = \frac{1\,\cdot\,3}{2}+\frac{1\,\cdot\,3}{2\,\cdot\,4}+\frac{1\,\cdot\,3\,\cdot\,5}{2\,\cdot\,4\,\cdot\,6}+\,\cdots\,+\frac{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)}{2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n)} \\[3mm] 
&\, \qquad = \frac{1\,\cdot\,3\,\cdot\,5}{2\,\cdot\,4}+\frac{1\,\cdot\,3\,\cdot\,5}{2\,\cdot\,4\,\cdot\,6}+\,\cdots\,+\frac{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)}{2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n)} \\[3mm] 
&\, \qquad = \,\cdots\, = \frac{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n+1)}{2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n)} = \color{red}{\frac{(2n+1)!!}{(2n)!!}} \,\Rightarrow\, S_{\small n}=\frac{(2n+1)!!}{(2n)!!}-1 \\[6mm] 
&\, 1+\lim_{n\rightarrow\infty}S_{\small n} = \lim_{n\rightarrow\infty}\frac{(2n+1)!!}{(2n)!!}\color{red}{\,\longrightarrow\,\infty} \qquad \{{\small\text{divergent}}\}
\end{align} 
$$
A: You are dealing with
$$ \sum_{k=1}^{n}\frac{(2k-1)!!}{(2k)!!} = \sum_{k=1}^{n}\frac{1}{4^k}\binom{2k}{k}.\tag{1}$$
It is well-known that $\frac{1}{4^k}\binom{2k}{k}\sim\frac{1}{\sqrt{\pi k}}$ as $k\to +\infty$, hence the associated series is divergent by the p-test. We may also notice that, by the extended binomial theorem,
$$ \sum_{k\geq 1}\frac{x^k}{4^k}\binom{2k}{k} = \frac{1}{\sqrt{1-x}}-1 \tag{2}$$
hence the RHS of $(1)$ is the coefficient of $x^n$ in the Taylor series (at $x=0$) of
$$ \frac{1}{1-x}\left(\frac{1}{\sqrt{1-x}}-1 \right) = \frac{1}{(1-x)^{3/2}}-\frac{1}{1-x}\tag{3} $$
and by the extended binomial theorem (again):
$$ \sum_{k=1}^{n}\frac{1}{4^k}\binom{2k}{k} = \color{red}{\frac{2n+1}{4^n}\binom{2n}{n}-1}.\tag{4} $$
Obviously $(4)$ can be proved through creative telescoping or probabilistic arguments, too. Interesting question.
A: Hint
$$\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot ...\cdot \frac{2n-1}{2n}= \frac{(2n-1)!}{2^n n!}$$
A: If $x=-1,n=-1/2$ then your series is $$nx+\frac{n(n-1)}{2!}x^{2}+\cdots$$ which is the general binomial series for $(1+x)^{n}$ except for the missing first term $1$. It is known from theory of binomial series (see towards the end of this blog post) that if $x=-1$ then the binomial series is convergent only when $n\geq 0$. Thus the given series diverges. 
