how to evaluate $\lim_{n \to \infty} n\frac{(n-1)!!}{n!!} $

Question

I want to evaluate $\lim_{n \to \infty} n\frac{(n-1)!!}{n!!} $

I was solving the integral:

$$I_n:=\lim_{ n \to \infty} \int_0 ^n \left( \frac{2nt}{t^2+n^2} \right)^n dt$$

and I did the following let $t=n \tan(x)$ , $dt= n \sec^2{(x)} dx$then $$I_n =\lim_{n \to \infty} n\int_0 ^{\frac{\pi}{4}} \sin^n{(2x)} \sec^2{(x)}dx =\lim_{n \to \infty} n\left( \int_0 ^{\frac{\pi}{4}} \sin^n{(2x)}dx +\int_0 ^{\frac{\pi}{4}} \sin^n{(2x)} \tan^2{(x)}dx \right) \geq \lim_{n \to \infty} \frac{n}{ 2} \int_0 ^{\frac{\pi}{2}} \sin^n{(x)}dx$$

There is famous formula (Wallis formula) for $\int_0 ^{\frac{\pi}{2}} \sin^n{(x)}dx$ which is $\frac{(n-1)!!}{n!!} $ if $n$ is odd and $ \frac{(n-1)!!}{n!!} \frac{ \pi}{2}$ if $n$ is even,I do believe the integral diverge to infinity i.e $\lim_{n \to \infty} n\frac{(n-1)!!}{n!!}= \infty $ but I couldn't prove that.

Answer 1

Define a sequence $(a_n)_{n\ge 1}$ by $a_n:=n\frac{(n-1)!!}{n!!}$. We find that for all $n\ge 1$, $$a_{n+2}=\left(1+\frac1n\right)a_n.$$ Now, $$a_{2n}=\frac{a_{2n}}{a_2}=\prod_{k=1}^{n-1}\left(1+\frac1{2k}\right)\ge\frac12\sum_{k=1}^{n-1}\frac1k,$$ which famously diverges as $n\to\infty$. Do the same for odd indices and you'll find $$\lim_{n\to\infty}a_n=\infty,$$ as desired.

Answer 2

Transforming the Wallis integral into a factorial and then use the Stirling formula (the genralised one on double factorials) is weird because we actually compute an equivalent of the Wallis integrals in order to prove the Stirling formula. In fact, it can be proven directly that $W_n \sim \sqrt{\frac{\pi}{2n}}$.

For this, use the induction formula (proven thanks to an integration by parts), $W_{n + 2} = \frac{n + 1}{n + 2}W_n$, and the fact that $(W_n)$ is positive decreasing so $\frac{n + 1}{n + 2} = \frac{W_{n + 2}}{W_n} \leqslant \frac{W_{n + 1}}{W_n} \leqslant 1$ hence $W_{n + 1} \sim W_n$.

We deduce that $W_n^2 \sim W_{n + 1}W_n$ and the sequence $u_n = (n + 1)W_{n + 1}W_n$ verifies the indeuciton relation $u_{n + 1} = (n + 2)W_{n + 1}W_{n + 2} = (n + 1)W_{n + 1}W_n$ therefore it is constant and it equals $u_0 = W_1W_0 = \frac{\pi}{2}$ so $W_n^2 \sim \frac{u_n}{n + 1} \sim \frac{\pi}{2n}$. We deduce that indeed $W_n \sim \sqrt{\frac{\pi}{2n}}$.

From this, it is clear that $\frac{n}{2}W_n \sim \sqrt{\frac{\pi n}{8}} \rightarrow +\infty$.

By the way, (if your change of variable is correct, I didn't check), $\displaystyle I_n = n\int_0^{\pi/4} \frac{\sin^n(2x)}{\cos^2(x)} \, dx \geqslant n\int_0^{\pi/4} \frac{\sin^n(2x)}{\cos^2(0)} \, dx = n\int_0^{\pi/4} \sin^n(2x) \, dx = \frac{n}{2}W_n \rightarrow +\infty$.

I think this last line is what you wanted to write but you wrote $\leq$ instead of $\geq$.

Answer 3

$$ n\frac{(n-1)!!}{n!!} = n \frac{(n-1)(n-3)\cdots}{n(n-2)(n-4)\cdots} = \frac{(n-1)!!}{(n-2)!!}. $$

$$ \ln\left(\frac{ (n-1)!! }{ (n-2)!! }\right) = \ln \left( \frac{n-1}{n-2} \cdot \frac{n-3}{n-4} \cdot \cdots\right) = \ln \left( \frac{n-1}{n-2} \right) + \ln \left( \frac{n-3}{n-4} \right) + \ldots $$

$$ = \ln \left( 1 + \frac{1}{n-2} \right) + \ln \left( 1 + \frac{1}{n-4} \right) + \ldots $$

$$ \geq \frac{1}{2}\left( \frac{1}{n-2} \right) + \frac{1}{2}\left( \frac{1}{n-4} \right) + \ldots = \frac{1}{2} \left( \frac{1}{n-2} + \frac{1}{n-4} + \ldots \right) $$

$$ = \frac{1}{2} \cdot \frac{1}{2} \cdot \sum_{k=1}^{ \left\lfloor \frac{n-1}{2} \right\rfloor } \frac{1}{k} \overset{n\to\infty}{\longrightarrow}\infty. $$

This shows that $ \ln\left(\frac{ (n-1)!! }{ (n-2)!! }\right) \overset{n\to\infty}{\longrightarrow}\infty, \implies \frac{ (n-1)!! }{ (n-2)!! } \overset{n\to\infty}{\longrightarrow}\infty.$