$\lim_{n\to\infty} \frac{n!!}{(n+1)!!}=0$ please check my proof for  $a_n = \frac {n!!}{(n+1)!!}$, prove that $\lim a_n = 0$
putting $a_n$ into logarithm, we get
$$\ln a_n=\sum_{k=1}^n \ln \frac k {k+1} = -\sum_{k=1}^n \ln (1+\frac1 k)$$
by taylor series,  
$$\ln(1+1)=\frac 1 1 - \frac 1 2 + \frac 1 3+...$$
$$\ln(1+\frac1 2)=\frac 1 2 - \frac 1 8 + \frac 1 {24}-...$$
$$...$$
$$\ln(1+\frac1 k)=\frac 1 k - \frac 1 {2k^2} + \frac 1 {3k^3}-...$$
if "$\lim \ln a_n$" exists, $\sum_1^\infty \ln(1+\frac 1 k)$  (absolutely) converges. so we know that any rearrangement series of $\ln(1+\frac 1 k)$ converges and has the same value.
furthermore, $\ln(1+\frac 1 k) =\sum_1^\infty\frac {(-1)^{k+1}} k$ converges absolutely if $k>1$
so, by assuning $\ln a_n$ converges, expand $\ln(1+\frac 1 k)$ by taylor series for $k>1$ and rearrange column by column.
$$\sum_{k=1}^{\infty} \ln (1+\frac1 k)=(\ln2 + (\frac1 2 + \frac 1 3 + ...) - \frac 1 2 (\frac 1 4 + \frac 1 9 + ...) +\frac1 3(\frac 1 8 +\frac 1 {27} +...)+...)$$
however the second term $(\frac 1 2 + \frac 1 3 +...)$ doesn't converge while all the other terms converge. this contradiction proves that $\ln a_n$ doesn't converge. so we have
$$\lim_{n\to\infty}\ln a_n=-\sum_{k=1}^{\infty} \ln (1+\frac1 k)=-\infty$$
which implies $\lim a_n =0$.


*

*Is this proof correct?   

*Can you give me more simple proof?

 A: Your formula for $a_n$ is wrong,
because
$m!! = \prod_{k=0}^{\lfloor (m-1)/2 \rfloor} (m-2k)
$.
Therefore
$a_n
=\dfrac{n!!}{(n+1)!!}
=\dfrac{\prod_{k=0}^{\lfloor (n-1)/2 \rfloor} (n-2k)}{\prod_{k=0}^{\lfloor /2 \rfloor} (n+1-2k)}
$.
I'll show for even $n$
and leave odd $n$ up to you.
$\begin{array}\\
a_{2n}
&=\dfrac{(2n)!!}{(2n+1)!!}\\
&=\dfrac{\prod_{k=0}^{n-1} (2n-2k)}{\prod_{k=0}^{n} (2n+1-2k)}\\
&=\prod_{k=0}^{n-1} \left(\dfrac{2n-2k}{2n+1-2k}\right)\\
&=\prod_{k=0}^{n-1} \left(1-\dfrac{1}{2n+1-2k}\right)\\
&=\prod_{k=1}^{n} \left(1-\dfrac{1}{2k+1}\right)
\qquad\text{ putting } n-k \text{ for } k\\
\end{array}
$
Since
$\sum \dfrac{1}{2k+1}$
diverges,
the product goes to zero.
A: Note that for the odd terms, we have
$$a_{2n-1}=\frac{(2n-1)!!}{(2n)!!}=\frac{(2n)!}{4^n\,(n!)^2}$$
Using Stirling's Formula, we have
$$\begin{align}
a_{2n-1}\sim \frac{\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}}{4^n(2\pi n)\left(\frac{n}{e}\right)^{2n}}=\sqrt{\frac{\pi}{n}}
\end{align}$$
Therefore, $\lim_{n\to \infty}a_{2n-1}=0$.

We also have for the even terms
$$\begin{align}
a_{2n}&=\frac{(2n)!!}{(2n+1)!!}\\\\
&=\frac{1}{(2n+1)a_{2n-1}}\\\\
&\sim\frac{\sqrt{n}}{\sqrt{\pi}(2n+1)}\to 0
\end{align}$$
