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Let $n\in \mathbb{N}^{*}$, let ${\displaystyle u_{n}={2n \choose n}\sqrt{n}\ 4^{-n}}$

Show that $(u_{n})_{n}$ is convergent and ${l.e^{-\frac{1}{8n}}<u_{n}<l}$

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i'm trying to compute $\dfrac{u_{n+1}}{u_n}$ but i failed even with wolframe 

{u(n) = (binomial(2 n, n) sqrt(n))×4^(-n), (u(n+1))/(u(n))=? }

any help would be appreciated

Jack D'Aurizio
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Educ
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1 Answers1

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We have:

$$ u_n = \sqrt{n}\frac{(2n)!}{4^n n!^2} = \sqrt{n}\frac{(2n-1)!!}{(2n)!!} = \sqrt{n}\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right),$$ so: $$ u_n^2 = \frac{n}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}+\frac{1}{4k^2}\right)=\frac{n}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right) $$ or just : $$ u_n^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right). $$ Since the product, as $n\to +\infty$, is convergent, we have: $$ u_n^2 \leq \frac{1}{4}\prod_{k=1}^{+\infty}\left(1+\frac{1}{4k(k+1)}\right)=C.$$ On the other hand: $$ u_n^2 = \frac{C}{\prod_{k=n}^{+\infty}\left(1+\frac{1}{4k(k+1)}\right)}\geq\frac{C}{\exp\left(\frac{1}{4}\sum_{k=n}^{+\infty}\left(\frac{1}{k}-\frac{1}{k+1}\right)\right)}=C e^{-\frac{1}{4n}}. $$

Jack D'Aurizio
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  • what does !! means and how did you get $u_n = \sqrt{n}\frac{(2n)!}{4^n n!^2} = \sqrt{n}\frac{(2n-1)!!}{(2n)!!} = \sqrt{n}\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right),$ could you please elaborate – Educ Dec 14 '14 at 08:25
  • $$(2n)!!=(2n)(2n-2)\cdot\ldots\cdot 2$$ and $$(2n-1)!!=(2n-1)(2n-3)\cdot\ldots\cdot 1.$$ – Jack D'Aurizio Dec 14 '14 at 17:17
  • could you add another way please – Educ Dec 14 '14 at 18:34
  • @Educ: I do not have another way. – Jack D'Aurizio Dec 14 '14 at 19:17
  • if we have that $u_n^2 \geq C e^{-\frac{1}{4n}}$ what does mean please add more detaile to see why you prove this $u_n^2 \geq C e^{-\frac{1}{4n}}$ and how we can ${l.e^{-\frac{1}{8n}}<u_{n}<l}$ – Educ Dec 23 '14 at 18:52
  • @Educ: just take the square root. $l=\sqrt{C}$. – Jack D'Aurizio Dec 23 '14 at 20:57
  • @Educ: are you kidding me? The square root of $C e^{-\frac{1}{4n}}$ is $l e^{-\frac{1}{8n}}$ that is exactly what you need. How about studying basic calculus a little more before approaching this kind of subtle exercises? – Jack D'Aurizio Dec 23 '14 at 23:13
  • if you just could tell me the principale of what you did . you show that ^$(u_n)^2$ is convergent is that so we can say $u_n$ is convergent too, thanks for ur time – Educ Dec 24 '14 at 08:13
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    @Educ: sorry for being a bit harsh before. Yes, since $u_n^2$ is between $C e^{-\frac{1}{4n}}$ and $C$ and $u_n>0$, by taking the square root we have that $u_n$ is between $l,e^{-\frac{1}{8n}}$ and $l$ as wanted. – Jack D'Aurizio Dec 24 '14 at 08:58
  • how could you prove that $u_n^2$ is increasing – Educ Jan 07 '15 at 21:21
  • @Educ: that is really trivial. For any $n\geq 2$, $$\frac{u_{n+1}^2}{u_n^2}=1+\frac{1}{4n(n+1)}>1.$$ – Jack D'Aurizio Jan 08 '15 at 10:21