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

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

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## Comments

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TopNewestuse sandwich theorem , it greatly helps here

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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? :)

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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 : \(\displaystyle \prod_{n=1}^{\infty} \frac{2n-1}{2n} = \sqrt{\frac{2}{\pi}}\)

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This is incorrect. Putting \(x=\pi/2\) gives \(\dfrac 2\pi=\prod\limits_{n=1}^\infty\dfrac{(2n)^2-1}{(2n)^2}\). So, we have \(\sqrt{\dfrac 2\pi}=\prod\limits_{n=1}^\infty\dfrac{\sqrt{(2n)^2-1}}{2n}\) which is obviously not the same as the one you wrote.

I guess you mistook the numerator \((2n)^2-1\) as \((2n-1)^2\).

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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 \(\displaystyle\lim_{n \rightarrow \infty} \frac{\displaystyle\prod_{r=0}^n{(2r+1)}}{\displaystyle\prod_{r=1}^{n+1}{2r}} \)

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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 , \(\displaystyle \lim_{n\to\infty}\prod_{r=1}^{n+1}2r=\prod_{r=1}^{\infty}2r\)

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I used

wolfram alphafor 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.

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Legendre's duplication formula to get \(2^{2n+2}\Gamma(n+\frac 32)\Gamma(n+2)=\sqrt{\pi}\cdot\Gamma(2n+3)\) and using this along with \(\dfrac{\Gamma(2n+3)}{\left(\Gamma(n+2)\right)^2}=\dfrac{(2n+2)!}{\left((n+1)!\right)^2}=\dbinom{2n+2}{n+1}\), you could rewrite the limit as,

Yes, this is correct. The limit is 0. However, instead of using W|A for calculating the limit, you could use\[L=\lim_{n\to\infty}\frac{\Gamma(n+3/2)}{\sqrt\pi\cdot\Gamma(n+2)}=\lim_{n\to\infty}\frac{\Gamma(2n+3)}{2^{2n+2}\cdot\left(\Gamma(n+2)\right)^2}=\lim_{n\to\infty}\frac{\binom{2n+2}{n+1}}{(2^{n+1})^2}\]

Note that this is the second limit you mentioned in your comment. Rather than concluding that the limit is 0 using a handwavy argument that the binomial coefficient is a decreasing function (I can't see an easy way to verify it), I think a formal way would be to use a corollary of Stirling's result that \(\dbinom{2n}n\sim\dfrac{4^n}{\sqrt{\pi n}}\) as \(n\to\infty\), so the limit becomes,

\[L=\lim_{n\to\infty}\frac{4^{n+1}}{\sqrt{\pi(n+1)}(2^{n+1})^2}=\frac 1{\sqrt\pi}\lim_{n\to\infty}\frac{1}{\sqrt{n+1}}=0\]

where the penultimate limit is easily evaluated noting that \(\sqrt{n+1}\to\infty\) as \(n\to\infty\), hence \(\frac{1}{\sqrt{n+1}}\to 0\) as \(n\to\infty\).

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Also you don't need to go to Legendre's formula to get to the function I mentioned. Just multiply the numerator and denominator with the denominator(i.e. product of even numbers) again to get that.

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