Find the infinite product $$\left(\dfrac{2}{1}\cdot\dfrac{2}{3}\cdot\dfrac{4}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{5}\cdot\dfrac{6}{7}\cdot\dfrac{8}{7}\cdots\right)$$
I solved it, but my method is not nice one. I solved it using calculus (that's why tagged). Please give a nice and sweet solution.
Answer
$\dfrac{\pi}{2}$
We can write the product as a limit of a ratio of gamma functions as follows $$ \eqalign{ & P = {2 \over 1}{2 \over 3}{4 \over 3}{4 \over 5}{6 \over 5}{6 \over 7} \cdots = \cr & = \prod\limits_{1\, \le \,n} {{{2n} \over {2n - 1}}{{2n} \over {2n + 1}}} = \cr & = \prod\limits_{1\, \le \,n} {{n \over {n - 1/2}}{n \over {n + 1/2}}} = \cr & = \prod\limits_{0\, \le \,k} {{{1 + k} \over {1/2 + k}}{{1 + k} \over {3/2 + k}}} = \cr & = \mathop {\lim }\limits_{m\; \to \;\infty } \prod\limits_{k = 0}^{m - 1} {{{1 + k} \over {1/2 + k}}{{1 + k} \over {3/2 + k}}} = \cr & = \mathop {\lim }\limits_{m\; \to \;\infty } {{1^{\,\overline {\,m} } } \over {\left( {1/2} \right)^{\,\overline {\,m} } }} {{1^{\,\overline {\,m} } } \over {\left( {3/2} \right)^{\,\overline {\,m} } }} = \cr & = \mathop {\lim }\limits_{m\; \to \;\infty } {{{{\Gamma \left( {1 + m} \right)} \over {\Gamma \left( 1 \right)}}} \over {{{\Gamma \left( {1/2 + m} \right)} \over {\Gamma \left( {1/2} \right)}}}}{{{{\Gamma \left( {1 + m} \right)} \over {\Gamma \left( 1 \right)}}} \over {{{\Gamma \left( {3/2 + m} \right)} \over {\Gamma \left( {3/2} \right)}}}} = \cr & = \mathop {\lim }\limits_{m\; \to \;\infty } {{{{\Gamma \left( {1 + m} \right)} \over {\Gamma \left( 1 \right)}}} \over {{{\Gamma \left( {1/2 + m} \right)} \over {\Gamma \left( {1/2} \right)}}}}{{{{\Gamma \left( {1 + m} \right)} \over {\Gamma \left( 1 \right)}}} \over {{{\Gamma \left( {3/2 + m} \right)} \over {\Gamma \left( {3/2} \right)}}}} = \cr & = {{\Gamma \left( {1/2} \right)} \over 1}{{\Gamma \left( {3/2} \right)} \over 1}\mathop {\lim }\limits_{m\; \to \;\infty } {{\Gamma \left( {1 + m} \right)} \over {\Gamma \left( {1/2 + m} \right)}}{{\Gamma \left( {1 + m} \right)} \over {\Gamma \left( {3/2 + m} \right)}} \cr} $$ where
$$ x^{\,\overline {\,m} } = \prod\limits_{k = 0}^{m - 1} {x + k} = {{\Gamma \left( {x + m} \right)} \over {\Gamma \left( x \right)}} $$ represents the Rising Factorial;
To the gamma functions we can apply the Stirling's approximation to get $$ \eqalign{ & P = \cr & = {{\Gamma \left( {1/2} \right)} \over 1}{{\Gamma \left( {3/2} \right)} \over 1}\mathop {\lim }\limits_{m\; \to \;\infty } {{\Gamma \left( {1 + m} \right)} \over {\Gamma \left( {1/2 + m} \right)}}{{\Gamma \left( {1 + m} \right)} \over {\Gamma \left( {3/2 + m} \right)}} = \cr & = {{\Gamma \left( {1/2} \right)^{\,2} } \over 2}\mathop {\lim }\limits_{m\; \to \;\infty } {{\left( {\sqrt {\,{{2\,\pi } \over {1 + m}}\,} \left( {{{1 + m} \over e}} \right)^{\,1 + m} } \right)^{\,2} } \over {\sqrt {\,{{2\,\pi } \over {1/2 + m}}\,} \left( {{{1/2 + m} \over e}} \right)^{\,1/2 + m} \sqrt {\,{{2\,\pi } \over {3/2 + m}}\,} \left( {{{3/2 + m} \over e}} \right)^{\,m + 3/2} }} = \cr & = {\pi \over 2}\mathop {\lim }\limits_{m\; \to \;\infty } {{\sqrt {\left( {1/2 + m} \right)\left( {3/2 + m} \right)} } \over {\left( {m + 1} \right)}}{{\left( {\left( {1 + m} \right)^{\,1 + m} } \right)^{\,2} } \over {\left( {1/2 + m} \right)^{\,1/2 + m} \left( {3/2 + m} \right)^{\,m + 3/2} }} = \cr & = {\pi \over 2}\mathop {\lim }\limits_{m\; \to \;\infty } {{\left( {1 + m} \right)^{\,2m + 1} } \over {\left( {1/2 + m} \right)^{\,m} \left( {3/2 + m} \right)^{\,m + 1} }} = \cr & = {\pi \over 2}\mathop {\lim }\limits_{m\; \to \;\infty } {{\left( {1 + m} \right)m^{\,2m} } \over {\left( {3/2 + m} \right)m^{\,2m} }}{{\left( {1 + 1/m} \right)^{\,2m} } \over {\left( {1 + 1/\left( {2m} \right)} \right)^{\,m} \left( {1 + 1/\left( {2/3m} \right)} \right)^{\,m} }} = \cr & = {\pi \over 2}{{e^{\,2} } \over {e^{\,1/2} e^{\,3/2} }} = {\pi \over 2} \cr} $$