To show for following sequence $\lim_{n \to \infty} a_n = 0$ where $a_n$ = $1.3.5 ... (2n-1)\over 2.4.6...(2n)$

Question

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

Answer 1

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 $$a_n \approx \frac{1}{\sqrt{\pi n} }\Big(1-\frac{1}{8n}\Big)$$

Answer 2

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 $n\to\infty$.

Answer 3

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$$12n^3+28n^2+19n+4\le12n^3+28n^2+20n+4.$$

Answer 4

$\forall k\in\mathbb{N}$
$k^2+2k<k^2+2k+1\Rightarrow k(k+2)<(k+1)^2\Rightarrow\frac{k}{k+1}<\frac{k+1}{k+2}$

$\therefore a_n=\frac{1\cdot3\cdot5...(2n-1)}{2\cdot4\cdot6...2n}=\frac{1}{2}\cdot\frac{3}{4}...\frac{k}{k+1}...\frac{2n-1}{2n}<\frac{2}{3}\cdot\frac{4}{5}...\frac{k+1}{k+2}...\frac{2n}{2n+1}=\frac{2\cdot4\cdot6...2n}{1\cdot3\cdot5...(2n+1)}$
where, $k$ is an odd natural number.

$\Rightarrow {a_n}^2<\frac{1\cdot3\cdot5...(2n-1)}{2\cdot4\cdot6...2n}\cdot\frac{2\cdot4\cdot6...2n}{1\cdot3\cdot5...(2n+1)}=\frac{1}{2n+1}$

$\Rightarrow 0<a_n<\frac{1}{\sqrt{2n+1}}$

Using Squeeze theorem, we get $\underset{n\rightarrow\infty}{\lim} a_n=0$.

Answer 5

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 $0$.

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

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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.