Limit of a division of products

limnr=0n(2r+1)r=1n+12r\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 00. Please post your solution to this.

Note by Swagat Panda
3 years, 2 months ago

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use sandwich theorem , it greatly helps here

Ujjwal Mani Tripathi - 3 years, 1 month ago

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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? :)

Swagat Panda - 3 years, 1 month ago

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The infinite product representation of sinx\sin x is sinxx=r=1(1x2r2π2)\displaystyle \frac{\sin x}{x} = \prod_{r=1}^{\infty} (1-\frac{x^2}{r^2 \pi^2} )

Put x=π2\displaystyle x=\frac{\pi}{2} to check that , 2π=1.3.3.5.5.7.72.2.4.4.6.6=(1.3.52.4.6)2\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 : n=12n12n=2π\displaystyle \prod_{n=1}^{\infty} \frac{2n-1}{2n} = \sqrt{\frac{2}{\pi}}

Aditya Narayan Sharma - 3 years, 2 months ago

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Okay, but my question is different from your expression in the last step, you have solved for n=12n12n\displaystyle\prod_{n=1}^{\infty}{\frac{2n-1}{2n}} ,whereas I asked for limnr=0n(2r+1)r=1n+12r\displaystyle\lim_{n \rightarrow \infty} \frac{\displaystyle\prod_{r=0}^n{(2r+1)}}{\displaystyle\prod_{r=1}^{n+1}{2r}}

Swagat Panda - 3 years, 2 months ago

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Observe that r=0n(2r+1)=r=1n(2(r1)+1)=r=1n(2r1)\prod_{r=0}^{n} (2r+1)=\prod_{r=1}^{n}(2(r-1)+1)=\prod_{r=1}^{n}(2r-1) , Put limit to obtain limnr=0n(2r+1)=r=1(2n1)\displaystyle \lim_{n\to\infty}\prod_{r=0}^{n}(2r+1)=\prod_{r=1}^{\infty}(2n-1)

Similarly , limnr=1n+12r=r=12r\displaystyle \lim_{n\to\infty}\prod_{r=1}^{n+1}2r=\prod_{r=1}^{\infty}2r

Aditya Narayan Sharma - 3 years, 2 months ago

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@Aditya Narayan Sharma I think the answer is 00 because, limnr=0n(2r+1)r=1n+12r=limn135(2n+1)246(2(n+1))\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 π/2\pi/2, and using Walli's formula, we get: limnπ2135(2n+1)246(2(n+1))=limn0π/2sin2n+2x.dx=limnΓ(32+n)πΓ(2+n)=0limnr=0n(2r+1)r=1n+12r=2π×(limnπΓ(32+n)2Γ(2+n))=2π×0=0\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

limn(2n+2)!(2n+1(n+1)!)2=limn(2n+2n+1)(2n+1)2=0\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(2n+2n+1)(2n+1)2\dfrac {\dbinom{2n+2}{n+1} }{ (2^{ n+1 })^{ 2 } } is a decreasing function.

Swagat Panda - 3 years, 2 months ago

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@Swagat Panda Yes, this is correct. The limit is 0. However, instead of using W|A for calculating the limit, you could use Legendre's duplication formula to get 22n+2Γ(n+32)Γ(n+2)=πΓ(2n+3)2^{2n+2}\Gamma(n+\frac 32)\Gamma(n+2)=\sqrt{\pi}\cdot\Gamma(2n+3) and using this along with Γ(2n+3)(Γ(n+2))2=(2n+2)!((n+1)!)2=(2n+2n+1)\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,

L=limnΓ(n+3/2)πΓ(n+2)=limnΓ(2n+3)22n+2(Γ(n+2))2=limn(2n+2n+1)(2n+1)2L=\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 (2nn)4nπn\dbinom{2n}n\sim\dfrac{4^n}{\sqrt{\pi n}} as nn\to\infty, so the limit becomes,

L=limn4n+1π(n+1)(2n+1)2=1πlimn1n+1=0L=\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 n+1\sqrt{n+1}\to\infty as nn\to\infty, hence 1n+10\frac{1}{\sqrt{n+1}}\to 0 as nn\to\infty.

Prasun Biswas - 3 years ago

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@Prasun Biswas I asked for help because I couldn't find a way to interpret this limit formally using methods mentioned in the JEE syllabus and I was relatively new to Stirling's approximation then, which is why I gave the reason of decreasing function. Thanks for helping me out.

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.

Swagat Panda - 3 years ago

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This is incorrect. Putting x=π/2x=\pi/2 gives 2π=n=1(2n)21(2n)2\dfrac 2\pi=\prod\limits_{n=1}^\infty\dfrac{(2n)^2-1}{(2n)^2}. So, we have 2π=n=1(2n)212n\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)21(2n)^2-1 as (2n1)2(2n-1)^2.

Prasun Biswas - 3 years ago

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