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Finding value of $\displaystyle \lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{4k^2}{4k^2-1}$

Try: $$\lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{2k}{2k-1}\cdot \frac{2k}{2k+1} = \lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{2k}{2k-1}\cdot \prod^{n}_{k=1}\frac{2k}{2k+1}$$

$$\lim_{n\rightarrow \infty}\frac{(2\cdot 4 \cdot 6\cdots 2n)^2}{1\cdot 2\cdot 3\cdots 2n}\times \frac{(2\cdot 4 \cdot 6 \cdots 2n)^2}{1\cdot 2\cdot 3\cdots 2n}$$

$$\lim_{n\rightarrow \infty}\frac{(2\cdot 4\cdot 6\cdot 2n)^4}{(1\cdot 2\cdot 3\cdots 2n)^2} = \lim_{n\rightarrow \infty}2^{4n}\cdot \frac{(n!)^4}{(2n!)^2}$$

Did not find any clue how to solve from that point

Could some help me to solve it, Thanks

DXT
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2 Answers2

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With Stirling's approximation $n!\sim\sqrt{2\pi n}(n/e)^n$ we get: \begin{align} \prod^{n}_{k=1}\frac{4k^2}{4k^2-1} &=\prod^{n}_{k=1}\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}\\ &=\prod^{n}_{k=1}\frac{(2k)^2}{(2k-1)(2k)}\cdot\frac{(2k)^2}{(2k)(2k+1)}\\ &=4^n\cdot\prod^{n}_{k=1}\frac{k^2}{(2k-1)(2k)}\cdot 4^n\cdot\prod^{n}_{k=1}\frac{k^2}{(2k)(2k+1)}\\ &=2^{4n}\cdot\frac{(n!)^2}{(2n)!}\cdot\frac{(n!)^2}{(2n+1)!}\\ &=2^{4n}\cdot\frac{(n!)^4}{((2n)!)^2(2n+1)}\\ &\sim 2^{4n}\cdot\frac{(\sqrt{2\pi n}(n/e)^n)^4}{(\sqrt{2\pi 2n}(2n/e)^{2n})^2(2n+1)}\\ &\sim 2^{4n}\cdot\frac{4\pi^2n^2(n/e)^{4n}}{4\pi n(2n/e)^{4n}(2n)}\\ &\sim 2^{4n}\cdot\frac{\pi n^2}{2n^2 2^{4n}}\\ &\xrightarrow{n\to\infty}\frac\pi 2\\ \end{align}

Fabio Lucchini
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You may find the following approach useful which avoids Stirling'approximation.

Let $$a_n=\int_{0}^{\pi/2}\sin^nx\,dx\tag{1}$$ and using integration by parts we have $$a_n=\left.-\sin^{n-1}x\cos x\right|_{x=0}^{x=\pi/2}+(n-1)\int_{0}^{\pi/2}\sin^{n-2}x\cos^2x\,dx$$ and the last integral can be written as $a_{n-2}-a_n$ via the identity $\cos^2x=1-\sin^2x$ and thus we get the recurrence relation $$a_n=\frac{n-1}{n}a_{n-2}\tag{2}$$ Using the above relation repeatedly we get $$a_{2n}=\frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\dots\frac{1}{2}a_0\tag{3}$$ and $$a_{2n+1}=\frac{2n}{2n+1}\cdot \frac{2n-2}{2n-1}\dots \frac{2}{3}a_1\tag{4}$$ Noting that $a_0=\pi/2,a_1=1$ we have via $(3),(4)$ $$\frac{a_{2n}}{a_{2n+1}}=\frac{\pi} {2}\prod_{k=1}^{n}\frac{4k^2-1}{4k^2}$$ The LHS of the above equation tends to $1$ as shown later in this answer and hence the product in your question evaluates to $\pi/2$.

It is easy to observe that $$a_{2n+1}\leq a_{2n}\leq a_{2n-1}$$ and hence $$1\leq \frac{a_{2n}}{a_{2n+1}}\leq \frac{a_{2n-1}}{a_{2n+1}}=\frac{2n+1}{2n}$$ (via equation $(2)$). By squeeze theorem we see that $a_{2n}/a_{2n+1}\to 1$ as $n\to\infty $.

Note: For those not well acquainted with Stirling's formula, the argument in this answer is crucial in one of the proofs of Stirling's formula.

Paramanand Singh
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