Finding the limit of $\frac{(2n)!!}{(2n+1)!!}$ without using Stirling’s approximatation I have proven that the sequence is convergent.
Let $x_n=\frac{(2n)!!}{(2n+1)!!}$
$\frac{x_{n+1}}{x_n}$ $=\frac{2n+2}{2n+3}<1$ Therefore the sequence decreases. On the other hand
$x_n>0$. Therefore the sequence is bounded below.
Since we have a sequence that is bounded below and decreases the sequence must converge.
But from here I don’t know how to find the limit. Could you please help me? And could you please say if what I’ve done is necessary or not?
Thanks in advance!
 A: See that
$$x_n = \prod_{k=1}^n \frac{2k}{2k+1}, $$
so that
$$\ln x_n  = \sum_{k=1}^n \ln \left(\frac{2k}{2k+1}\right). $$
So we work with this series.  First argue that
$$\sum_{k=1}^{\infty} \ln\left(\frac{2k}{2k+1}\right) - \int_{t=1}^{\infty} \ln\left(\frac{2t}{2t+1}\right) \; dt$$
finite.  (If you graph the summmand and the integrand, they differ by a set of little (near) triangles.  Because the functions are monotonic, the triangles can be pushed together and fit in a small rectangle.)
Integration by parts gives
$$ = \int \ln(2t) - \ln(2t+1) \; dt =  t \ln(2t) -\frac{1}{2}\ln(2t+1)(2t+2) + 1/2.$$
And when you evaluate the improper integral you get $-\infty.$   So the limit of the sequence is $e^{-\infty} = 0.$
A: Observe that, for any $ n\in\mathbb{N} $ : $$\frac{\left(2n\right)!!}{\left(2n+1\right)!!}=\int_{0}^{1}{\left(1-x^{2}\right)^{n}\,\mathrm{d}x} $$
Since : \begin{aligned} \int_{0}^{1}{\left(1-x^{2}\right)^{n}\,\mathrm{d}x}&=\int_{0}^{1}{\left(1+x\right)^{n}\left(1-x\right)^{n}\,\mathrm{d}x}\\&\leq\int_{0}^{1}{\left(1+x\right)^{n}\,\mathrm{d}x}=\frac{1}{n+1}\underset{n\to +\infty}{\longrightarrow}0\end{aligned}
Then : $$ \frac{\left(2n\right)!!}{\left(2n+1\right)!!}\underset{n\to +\infty}{\longrightarrow}0 $$
A: Similar to another answer, you have that $$x_n = \prod_{k=1}^{n}\frac{2k}{2k+1}$$
Then
$$\ln(x_n) = \sum_{k=1}^{n}\ln\left( \frac{2k}{2k+1} \right)$$
Each term will be negative and with the comparison test, you have that $$\sum_{k=1}^{n}\ln\left( \frac{2k}{2k+1} \right) < -\sum_{k=1}^n \frac{1}{2k+1} \to -\infty$$
Then $\ln(x_n)$ must go to $-\infty$ as well, and $x_n = \exp(\ln(x_n)) \to 0$.
