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How can I show

$\lim_{n \to \infty} a_n = 0$

$a_n = {1.3.5 ... (2n-1)\over 2.4.6...(2n)}$

I have shown that $a_n$ is monotonically decreasing. I thought to shown sequence is bounded from below then it automatically would converge and hence my question will be solve. But I'm unable to show its boundedness... Or there maybe another method to prove this. Thanks

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  • $\begingroup$ Clarification request: what do you mean by $1.3.5...$? Are these numbers being added? $\endgroup$ – Mike Pierce Nov 17 '14 at 8:14
  • $\begingroup$ I think is multiplication of odd numbers. Isn't $\pi$ in the answer? I vaguely remember a result like this from Euler. $\endgroup$ – ReverseFlow Nov 17 '14 at 8:15
  • $\begingroup$ It is certainly bounded below, since all terms are positive, so it converges. But that does not show it converges to $0$. The Stirling approximation will show convergence to $0$, but one can do it with less machinery. $\endgroup$ – André Nicolas Nov 17 '14 at 8:21
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    $\begingroup$ Try to prove using induction that:$$a_n\lt\frac{1}{\sqrt{3n+1}}$$ $\endgroup$ – pointer Nov 17 '14 at 8:26
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    $\begingroup$ That the limit is zero is intuitively obvious, seeing as $a_n=\binom{2n}n/4^n$is the probability of getting exactly $n$ heads in $2n$ independent tosses of a fair coin. $\endgroup$ – bof Nov 17 '14 at 8:30
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Multiply the numerator by the denominator; so $$a_n=\frac{1\times3\times5\times ... \times(2n-1)}{ 2\times4\times6\times...\times(2n)}=\frac{1\times2\times3\times ... \times(2n)}{\Big(2\times4\times6\times...\times(2n)\Big)^2}=\frac{(2n)!}{4^n(n!)^2}$$ If now you use Stirling approximation $$m! \approx \sqrt{2 \pi } e^{-m} m^{m+\frac{1}{2}}$$ and then $$a_n \approx \frac{1}{\sqrt{\pi n} }$$ A more detailed approach would show for the asymptotic behavior $$a_n \approx \frac{1}{\sqrt{\pi n} }\Big(1-\frac{1}{8n}\Big)$$

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  • $\begingroup$ In first step how have u written in denominatorof last inequality as $4^{n}(n!)^{2}$ .. $\endgroup$ – godonichia Nov 17 '14 at 12:11
  • $\begingroup$ Factor $2$ inside the brackets of denominator and square. $\endgroup$ – Claude Leibovici Nov 17 '14 at 17:21
  • $\begingroup$ @ClaudeLeibovici in stirlings aproximstion as you have mentioned in order to find formula for 2m! I just have to replace m by 2m in above formula .if i do this then im not getting to answer, thanks $\endgroup$ – Jessica Gtb Dec 25 '14 at 12:20
  • $\begingroup$ @JessicaGtb. I don't know whee you had a problem since if you use Stirling, you should get $$(2n)! \approx \sqrt{\pi } 2^{2 m+1} e^{-2 m} m^{2 m+\frac{1}{2}}$$ Post if you still have problems. Cheers :-) $\endgroup$ – Claude Leibovici Dec 26 '14 at 7:22
  • $\begingroup$ How did you get that expression for 2n! .it is not same when you replace n by 2n in formula you mentioned in answer $\endgroup$ – Jessica Gtb Dec 26 '14 at 12:29
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Using $\boldsymbol{1+x\le e^x}$

Since $1+x\le e^x$ for all $x\in\mathbb{R}$, $$ \begin{align} a_n &=\prod_{k=1}^n\frac{2k-1}{2k}\\ &=\prod_{k=1}^n\left(1-\frac1{2k}\right)\\ &\le\prod_{k=1}^ne^{-\large\frac1{2k}}\\[3pt] &=e^{-\frac12H_n}\tag{1} \end{align} $$ where $H_n$ are the Harmonic Numbers. Since the Harmonic Series diverges, $(1)$ tends to $0$.


Using Gautschi's Inequality $$ \begin{align} a_n &=\prod_{k=1}^n\frac{2k-1}{2k}\\ &=\prod_{k=1}^n\frac{k-\frac12}{k}\\ &=\frac{\Gamma\!\left(n+\frac12\right)}{\sqrt\pi\,\Gamma(n+1)}\\[6pt] &\le\frac1{\sqrt{\pi n}}\tag{2} \end{align} $$


Commentary

Gautschi's Inequality also says that $$ \begin{align} a_n &=\frac{\Gamma\!\left(n+\frac12\right)}{\sqrt\pi\,\Gamma(n+1)}\\ &\ge\frac1{\sqrt{\pi\left(n+\frac12\right)}}\tag{3} \end{align} $$ Therefore, $$ \lim_{n\to\infty}\sqrt{n}\,a_n=\frac1{\sqrt\pi}\tag{4} $$ Putting the second line of $(1)$ and the limit in $(4)$ together, we can show that $$ \gamma+\sum_{n=2}^\infty\frac{\zeta(n)}{n2^{n-1}}=\log(\pi)\tag{5} $$ where $\gamma$ is the Euler-Mascheroni Constant and $\zeta$ is the Riemann Zeta Function.

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Let's prove using induction that $$a_n\le\frac{1}{\sqrt{3n+1}}.$$ For $n=1$ it is true. Now we just need to prove that$$\frac{(2n+1)^2}{(2n+2)^2}\le\frac{3n+1}{3n+4}$$or $$(4n^2+4n+1)(3n+4)\le(4n^2+8n+4)(3n+1)$$or$$12n^3+28n^2+19n+4\le12n^3+28n^2+20n+4.$$

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  • $\begingroup$ If we are not given the limit it converges to then how would we go about it $\endgroup$ – godonichia Nov 17 '14 at 8:51
  • $\begingroup$ I don't know. Maybe we could find asymtothics using Stirling's formula, then try to find this inequality. $\endgroup$ – pointer Nov 17 '14 at 9:08
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Your sequence can be rewritten as $$a_n=\frac{1\cdot3\cdot...\cdot(2n-1)}{2\cdot4\cdot...\cdot2n}=\frac{1\cdot2\cdot3\cdot...\cdot(2n-1)\cdot2n}{(2\cdot4\cdot...\cdot2n)^2}=\frac{(2n)!}{2^{2n}(n!)^2}$$ Using Stirling's approximation we get $$a_n=\frac{(2n)!}{2^{2n}(n!)^2}\sim\frac{1}{2^{2n}}\cdot\frac{\sqrt{2\pi 2n}\cdot(\frac{2n}{e})^{2n}}{(\sqrt{2\pi n}(\frac{n}{e})^n)^2}=\frac{1}{2^{2n}}\cdot\frac{\sqrt{2}\cdot2^{2n}}{\sqrt{2\pi n}}=\frac{\sqrt{2}}{\sqrt{2\pi n}}\to 0$$ as $n\to\infty$.

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  • $\begingroup$ i havent studied stirlings approximation yet . Can u show me another way which uses sequence series concepts $\endgroup$ – godonichia Nov 17 '14 at 8:35

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