Evaluate $\prod_{n\geq2}\left(\frac{4}{e^2}\left(1+\frac{1}{n}\right)^{2n+1}\frac{n^2-1}{4n^2-1}\right)$

Question

$$\prod_{n\geq2}\left(\frac{4}{e^2}\left(1+\frac{1}{n}\right)^{2n+1}\frac{n^2-1}{4n^2-1}\right)$$

I expand $\frac{n^2-1}{4n^2-1}$ as $\frac{(n-1)(n+1)}{(2n-1)(2n+1)}$. Then, $\left(1+\frac{1}{n}\right)^{2n+1}$ I expanded using the binomial theorem. $$\left(1+\frac{1}{n}\right)^{2n+1} = \sum_{k=0}^{2n+1}\binom{2n+1}{k}\left(\frac{1}{n}\right)^k$$ Still got stuck

Answer 1

Hint: $$\prod_{n\geq2}\left(\frac{4}{e^2}\left(1+\frac{1}{n}\right)^{2n+1}\frac{n^2-1}{4n^2-1}\right) \\\\ = \left({\prod \frac{4}{e^2} }\right)\left({\prod \frac{(n+1)^{n+1}}{n^n}}\right)^2\left({\prod \frac{n-1}{n}}\right)\prod \frac{1}{4n^2-1}$$

Answer 2

firstly by simplify we have $$ \prod_{n\geq 2} \left(e^{-2} \left(1+\frac{1}{n}\right)^{2n+1} \right) \prod_{n\geq 2} \frac{1-\frac{1}{n^2}}{1-\frac{1}{4n^2}} $$ using sine function we have $$ \prod_{n\geq 2} \left(1-\frac{x^2}{k^2} \right)=\frac{\sin(\pi x)}{\pi x(1-x^2)} $$ So $$ \prod_{n\geq 2} \frac{1-\frac{1}{n^2}}{1-\frac{1}{4n^2}}=\frac{3\pi}{16} $$ for other product use exponential function which known $" \exp(\ln x)=x"$ So, $$ \prod_{n\geq 2} \left(e^{-2} \left(1+\frac{1}{n}\right)^{2n+1} \right)=\exp\left(\ln \prod_{n\geq 2} \left(e^{-2} \left(1+\frac{1}{n}\right)^{2n+1} \right)\right)=\exp(\Omega)$$ and because of $$ \ln \left(\prod_{n\geq 2} f(k)\right)=\lim_{n\to \infty} \sum_{k=2}^n \ln f(k) $$ So $$ \Omega=\lim_{n\to \infty} \sum_{k=2}^n \left((2k+1) \ln \left(\frac{k+1}{k} \right)-2 \right) $$ $$ =\lim_{n\to \infty} \left(2\sum_{k=2}^n \left(k\ln(k+1)-k\ln k\right) +\ln(n+1)-\ln2-2(n-1)\right) $$ $$ =\lim_{n\to \infty} \left(2\sum_{k=2}^n \left((k-1)\ln k-k\ln k\right)+2n\ln(n+1)-2\ln2 +\ln(n+1)-2n\right)-\ln2+2 $$ $$ =\lim_{n\to \infty} \left(-2 \ln(n!)+(2n+1)\ln(n+1)-2n\right)-3\ln2+2 $$ and using Stirling's approximation which known $$ 2\ln(n!) \approx \ln(2\pi) +\left(2n+1 \right) \ln n-2n $$ So we can get easily $$ \Omega=4-\ln\pi-4\ln2 $$ then $$ \prod_{n\geq 2} \left(e^{-2} \left(1+\frac{1}{n}\right)^{2n+1} \right)=\frac{1}{16\pi}e^4 $$ finally $$ \prod_{n\geq 2} \left(\frac{4}{e^2} \left(1+\frac{1}{n}\right)^{2n+1} \frac{n^2-1}{4n^2-1} \right)=\frac{3}{256} e^4 $$

Answer 3

Stirling 's formula will be used $$\displaystyle{\mathop {\lim}\limits_{n \to \infty} \frac{{{e^n}\cdot n!}}{{\sqrt{2\pi n}\cdot {n^n} }} = 1}$$

$$\displaystyle{\prod\limits_{n = 2}^\infty {\left( {\frac{4}{{{e^2}}}{{\left( {1 + \frac{1}{n}} \right)}^{2n + 1}}\frac{{{n^2} - 1}}{{4{n^2} - 1}}} \right)} = \prod\limits_{n = 2}^\infty {\left( {\frac{1}{{{e^2}}}{{\left( {1 + \frac{1}{n}} \right)}^{2n + 1}}} \right)} \cdot \prod\limits_{n = 2}^\infty {\left( {\frac{{{n^2} - 1}}{{{n^2} - 1/4}}} \right)} }$$. Then

$$\displaystyle{\prod\limits_{n = 2}^\infty {\frac{{{n^2} - 1}}{{{n^2} - 1/4}}} = \mathop {\lim }\limits_{N \to \infty } \prod\limits_{n = 2}^N {\frac{{\left( {n - 1} \right)\left( {n + 1} \right)}}{{\left( {n - \frac{1}{2}} \right)\left( {n + \frac{1}{2}} \right)}}} = \mathop {\lim }\limits_{N \to \infty } \frac{{\left( {1 \cdot 2 \cdot 3 \cdot .. \cdot \left( {N - 1} \right)} \right) \cdot \left( {3 \cdot 4 \cdot 5 \cdot .. \cdot \left( {N + 1} \right)} \right)}}{{\left( {\frac{3}{2} \cdot \frac{5}{2} \cdot .. \cdot \frac{{2N - 1}}{2}} \right)\left( {\frac{5}{2} \cdot \frac{7}{2} \cdot .. \cdot \frac{{2N - 1}}{2} \cdot \frac{{2N + 1}}{2}} \right)}} = }$$

$$\displaystyle{ = \mathop {\lim }\limits_{N \to \infty } \frac{{3 \cdot \left( {2N + 1} \right) \cdot {2^{2\left( {N - 1} \right)}}}}{{2 \cdot N}} \cdot \frac{{{{\left( {{{\left( {N!} \right)}^2} \cdot \left( {N + 1} \right)} \right)}^2}}}{{{{\left( {3 \cdot 5 \cdot .. \cdot \left( {2N - 1} \right)\left( {2N + 1} \right)} \right)}^2}}} = \mathop {\lim }\limits_{N \to \infty } \left( {\frac{{3 \cdot \left( {2N + 1} \right) \cdot {2^{2\left( {N - 1} \right)}}}}{{2 \cdot N}} \cdot } \right.}$$

$$\displaystyle{\left. { \cdot \frac{{{{\left( {N!} \right)}^2}\left( {N + 1} \right){{\left( {{2^N}N!} \right)}^2}}}{{{{\left( {2 \cdot 3 \cdot 4 \cdot .. \cdot \left( {2N - 1} \right)\left( {2N} \right)\left( {2N + 1} \right)} \right)}^2}}}} \right) = \mathop {\lim }\limits_{N \to \infty } \frac{3}{8}\frac{{\left( {2N + 1} \right)\left( {N + 1} \right)}}{N} \cdot \frac{{{{\left( {N!} \right)}^4}{2^{4N}}}}{{{{\left( {\left( {2N + 1} \right)!} \right)}^2}}} = \mathop {.........}\limits^{Stirling} = \frac{{3\pi }}{{16}}}$$

and

$$\displaystyle{\prod\limits_{n = 2}^\infty {\left( {\frac{1}{{{e^2}}}{{\left( {1 + \frac{1}{n}} \right)}^{2n + 1}}} \right)} = \mathop {\lim }\limits_{N \to \infty } \prod\limits_{n = 2}^N {\left( {\frac{1}{{{e^2}}}{{\left( {1 + \frac{1}{n}} \right)}^{2n + 1}}} \right)} = \mathop {\lim }\limits_{N \to \infty } \frac{1}{{{e^{2N - 2}}}}\prod\limits_{n = 2}^N {{{\left( {\frac{{n + 1}}{n}} \right)}^{2n}}} \prod\limits_{n = 2}^N {\left( {\frac{{n + 1}}{n}} \right)} = }$$

$$\displaystyle{ = \mathop {\lim }\limits_{N \to \infty } \frac{1}{{{e^{2N - 2}}}}\frac{{{{\left( {N + 1} \right)}^{2N}}}}{{4{{\left( {N!} \right)}^2}}} \cdot \frac{{N + 1}}{2} = \mathop {...........}\limits^{Stirling} = \frac{{{e^4}}}{{16\pi }}}$$

Finally $$\displaystyle{\prod\limits_{n = 2}^\infty {\left( {\frac{4}{{{e^2}}}{{\left( {1 + \frac{1}{n}} \right)}^{2n + 1}}\frac{{{n^2} - 1}}{{4{n^2} - 1}}} \right)} = \frac{{3\pi }}{{16}} \cdot \frac{{{e^4}}}{{16\pi }} = \frac{{3{e^4}}}{{256}}}$$