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Limit of a division of products

\[\large\lim_{n \to\infty} \dfrac{\displaystyle\prod_{r=0}^n{(2r+1)}}{\displaystyle\prod_{r=1}^{n+1}{2r}} \]

I am unable to compute this limit mathematically, but my guess is that it is \(0\). Please post your solution to this.

Note by Swagat Panda
1 month, 2 weeks ago

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use sandwich theorem , it greatly helps here Starwar Clone · 1 month ago

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@Starwar Clone Thanks very much for the suggestion, but I don't know how to apply sandwich theorem in problems, I have only studied it in theory, can you give me a hint? :) Swagat Panda · 1 month ago

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The infinite product representation of \(\sin x\) is \(\displaystyle \frac{\sin x}{x} = \prod_{r=1}^{\infty} (1-\frac{x^2}{r^2 \pi^2} ) \)

Put \(\displaystyle x=\frac{\pi}{2}\) to check that , \(\displaystyle \frac{2}{\pi}=\frac{1.3.3.5.5.7.7\cdots}{2.2.4.4.6.6\cdots} = (\frac{1.3.5\cdots}{2.4.6\cdots})^2 \)

So your limit is : \(\displaystyle \prod_{n=1}^{\infty} \frac{2n-1}{2n} = \sqrt{\frac{2}{\pi}}\) Aditya Sharma · 1 month, 2 weeks ago

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@Aditya Sharma Okay, but my question is different from your expression in the last step, you have solved for \(\displaystyle\prod_{n=1}^{\infty}{\frac{2n-1}{2n}}\) ,whereas I asked for \(\displaystyle\lim_{n \rightarrow \infty} \frac{\displaystyle\prod_{r=0}^n{(2r+1)}}{\displaystyle\prod_{r=1}^{n+1}{2r}} \) Swagat Panda · 1 month, 2 weeks ago

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@Swagat Panda Observe that \(\prod_{r=0}^{n} (2r+1)=\prod_{r=1}^{n}(2(r-1)+1)=\prod_{r=1}^{n}(2r-1)\) , Put limit to obtain \(\displaystyle \lim_{n\to\infty}\prod_{r=0}^{n}(2r+1)=\prod_{r=1}^{\infty}(2n-1)\)

Similarly , \(\displaystyle \lim_{n\to\infty}\prod_{r=1}^{n+1}2r=\prod_{r=1}^{\infty}2r\) Aditya Sharma · 1 month, 2 weeks ago

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@Aditya Sharma I think the answer is \(0\) because, \(\displaystyle\lim _{ n\rightarrow \infty } \dfrac {\displaystyle \prod _{ r=0 }^{ n }{ (2r+1) } }{ \displaystyle\prod _{ r=1 }^{ n+1 }{ 2r } } =\displaystyle\lim _{ n\rightarrow \infty } \dfrac { 1\cdot 3\cdot 5\cdots (2n+1) }{ 2\cdot 4\cdot 6\cdots (2(n+1)) } \) Multiplying both sides by \(\pi/2\), and using Walli's formula, we get: \(\displaystyle\lim _{ n\rightarrow \infty } \dfrac\pi2\cdot\dfrac { 1\cdot 3\cdot 5\cdots (2n+1) }{ 2\cdot 4\cdot 6\cdots (2(n+1)) }=\displaystyle\lim _{ n\rightarrow \infty } \displaystyle \int_{0}^{{\pi}/{2}} {\sin ^{ 2n+2 }{ x }.dx }=\displaystyle\lim _{ n\rightarrow \infty }\dfrac{\Gamma \left({\dfrac32+n}\right)}{\sqrt π\cdot \Gamma{\left(2+n\right) }} =\boxed0 \\ \Rightarrow \displaystyle\lim _{ n\rightarrow \infty } \dfrac {\displaystyle \prod _{ r=0 }^{ n }{ (2r+1) } }{ \displaystyle\prod _{ r=1 }^{ n+1 }{ 2r } }=\dfrac2\pi \times\left(\displaystyle\lim _{ n\rightarrow \infty }\dfrac{\sqrt π \cdot \Gamma \left({\dfrac32+n}\right)}{2\cdot \Gamma{\left(2+n\right) }}\right)=\dfrac2\pi \times 0 =\boxed0 \)

I used wolfram alpha for computing the second last step, because I don't know how to compute limits in gamma functions.

The given function can also be written as

\[\displaystyle\lim_{n \rightarrow \infty}\dfrac{(2n+2)!}{\left(2^{n+1}\cdot(n+1)!\right)^{2}}=\displaystyle\lim_{n \rightarrow \infty}\dfrac {\dbinom{2n+2}{n+1} } { (2^{ n+1 })^{ 2 } }=0 \] because\(\dfrac {\dbinom{2n+2}{n+1} }{ (2^{ n+1 })^{ 2 } }\) is a decreasing function. Swagat Panda · 1 month, 1 week ago

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