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

Clarification request: what do you mean by $1.3.5...$? Are these numbers being added? — Mike Pierce, Nov 17 '14 at 08:14
I think is multiplication of odd numbers. Isn't $\pi$ in the answer? I vaguely remember a result like this from Euler. — ReverseFlowControl, Nov 17 '14 at 08:15
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. — André Nicolas, Nov 17 '14 at 08:21
Try to prove using induction that:$$a_n\lt\frac{1}{\sqrt{3n+1}}$$ — pointer, Nov 17 '14 at 08:26
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. — bof, Nov 17 '14 at 08:30
@user121270 Can u pls justify ur comment . Why have u choosen that specific thing ? — godonichia, Nov 17 '14 at 08:31
@AndréNicolas I havent studies stirlings approximation . If u know this by other means pls let me know — godonichia, Nov 17 '14 at 08:32
This limit is somewhat related: http://math.stackexchange.com/questions/1024290/how-to-evaluate-lim-n-to-infty-sqrtn-frac1-cdot3-cdot5-cdot-ldots-cdot2 — Martin Sleziak, Nov 17 '14 at 10:13
This question is very close to the OP: http://math.stackexchange.com/questions/584456/limit-of-a-fraction-of-double-factorials — Martin Sleziak, Nov 17 '14 at 10:36

reyna · Answer 1 · 2021-04-21T13:59:00.797

16

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

$\therefore a_n=\frac{1\cdot3\cdot5...(2n-1)}{2\cdot4\cdot6...2n}=\frac{1}{2}\cdot\frac{3}{4}...\frac{k}{k+1}...\frac{2n-1}{2n}<\frac{2}{3}\cdot\frac{4}{5}...\frac{k+1}{k+2}...\frac{2n}{2n+1}=\frac{2\cdot4\cdot6...2n}{1\cdot3\cdot5...(2n+1)}$
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 $\underset{n\rightarrow\infty}{\lim} a_n=0$.

edited Apr 21 '21 at 13:59

answered Nov 15 '19 at 10:33

reyna

2,101

1

I know this is a dumb question, but can you please tell me how did you infer $\therefore a_n=\frac{1\cdot3\cdot5...(2n-1)}{2\cdot4\cdot6...2n}<\frac{2\cdot4\cdot6...2n}{1\cdot3\cdot5...(2n+1)}$ from $n(n+2)<(n+1)^2\Rightarrow\frac{n}{n+1}<\frac{n+1}{n+2}$? Thanks!! – Natasha J Apr 21 '21 at 02:53
@NatashaJ I hope you have understood how we get $\frac{k}{k+1}<\frac{k+1}{k+2}$. Now check that, for $n=1$ you get $\frac{1}{2}<\frac{2}{3}$, for $n=3$ you get $\frac{3}{4}<\frac{4}{5}$, and so on. In other words for an odd $k$, you see that $\frac{k}{k+1}<\frac{k+1}{k+2}$. – reyna Apr 21 '21 at 13:54
@NatashaJ I have modified the step, hope it clears your doubt. – reyna Apr 21 '21 at 13:59
1

thanks so much. Now I get it! – Natasha J Apr 21 '21 at 15:00

score 8 · Answer 2 · edited Apr 13 '17 at 12:21

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.

score 5 · Answer 3 · answered Nov 17 '14 at 08:46

5

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

answered Nov 17 '14 at 08:46

pointer

1,811

If we are not given the limit it converges to then how would we go about it – godonichia Nov 17 '14 at 08:51
I don't know. Maybe we could find asymtothics using Stirling's formula, then try to find this inequality. – pointer Nov 17 '14 at 09:08
(+1) It's neater with $n - 1$ in place of $n$: $$ \left(\frac{2n - 1}{2n}\right)^2 \leqslant \frac{3n - 2}{3n + 1} \iff \frac{4n - 1}{4n^2} \geqslant \frac3{3n + 1} \iff (4n - 1)(3n + 1) \geqslant 12n^2 \iff n \geqslant 1. $$ – Calum Gilhooley Mar 24 '22 at 21:30

score 4 · Accepted Answer · answered Nov 17 '14 at 08:36

4

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

answered Nov 17 '14 at 08:36

Claude Leibovici

260,315

In first step how have u written in denominatorof last inequality as $4^{n}(n!)^{2}$ .. – godonichia Nov 17 '14 at 12:11
Factor $2$ inside the brackets of denominator and square. – Claude Leibovici Nov 17 '14 at 17:21
@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 – Jessica Gtb Dec 25 '14 at 12:20
@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 :-) – Claude Leibovici Dec 26 '14 at 07:22
How did you get that expression for 2n! .it is not same when you replace n by 2n in formula you mentioned in answer – Jessica Gtb Dec 26 '14 at 12:29
It is the same (just expanded accordingly). – Claude Leibovici Dec 26 '14 at 16:17

score 2 · Answer 5 · answered Nov 17 '14 at 08:34

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 $n\to\infty$.

i havent studied stirlings approximation yet . Can u show me another way which uses sequence series concepts — godonichia, Nov 17 '14 at 08:35

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

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