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I want to show that $$0<\lim_{n\to\infty}\frac{2^{n}n!}{n^{1/2}(2n-1)!!}<\infty$$ By induction, the sequence is bounded above by $2$ and decreasing, so by monotone convergence, it has a finite limit. I'm stuck showing that the limit is positive. Plugging in big numbers makes me guess it converges to $\sqrt{\pi}$, but I'm not sure how to show it is bounded below by a positive number.

user363464
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3 Answers3

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We have: $$ \frac{2^n n!}{\sqrt{n}(2n-1)!!} = \frac{1}{\sqrt{n}}\prod_{k=1}^{n}\frac{1}{1-\frac{1}{2k}}$$ whose square is given by $$ \frac{\frac{1}{n}}{\frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{2k}\right)^2}=\frac{4}{n\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)} $$ that boils down to: $$ 4\prod_{k=2}^{n}\left(1-\frac{1}{(2k-1)^2}\right). $$ Since $\sum_{k\geq 2}\frac{1}{(2k-1)^2}$ is convergent, the limit as $n\to +\infty$ of the previous product is a finite positive number. This proves the claim. Which number is given by the Weierstrass product for the cosine function, for instance.

As an alternative approach, it is enough to write $\frac{2^n n!}{\sqrt{n}(2n-1)!!}$ in terms of the $\Gamma$ function and exploit Gautschi's inequality. That instantly gives the value of the limit, too.

Jack D'Aurizio
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  • I sort of flirted with this idea on my own, except that I wasn't able to reach the conclusion that the product converges to a positive number. What does the convergence of the series have to do with it? – user363464 May 23 '17 at 19:16
  • @user363464: it is the very definition of convergent product. $\prod_{n\geq 1}(1+a_n)$ is convergent iff $\sum_{n\geq 1}a_n$ is convergent. – Jack D'Aurizio May 23 '17 at 19:57
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    the definition of convergent product that I had in mind was that the sequence of finite products converge, but I see that your comment is a theorem in my textbook (although much further ahead of the part I'm working on). – user363464 May 23 '17 at 20:05
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Since $\left( 2n-1\right)!!=\frac{\left( 2n\right)!}{n!2^n}$, you seek $$\lim_{n\to\infty}\frac{n!^24^n}{\left( 2n\right)!\sqrt{n}}=\lim_{n\to\infty}\frac{2\pi n^{2n+1}4^ne^{-2n}}{2^{2n+1/2}\sqrt{2\pi}n^{2n+1}e^{-2n}}=\sqrt{\pi}$$(as you guessed), using Stirling's approximation $n!\approx\sqrt{2\pi}n^{n+1/2}e^{-n}$. The ratio of these expressions is $\approx e^{1/\left( 12n\right)}$, so the error $\to 0$ as $n\to\infty$.

J.G.
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The ratio test gives

$$\lim_{+\infty} \frac {a_{n+1}}{a_n}=$$

$$2\lim_{+\infty}\frac {(2n-1)!!}{(2n+1)!!}=0 $$

for enough great $n $,

$$\frac {a_{n+1}}{a_n}<k <1$$

thus $(a_n) $ converges to 0.

hamam_Abdallah
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