Is there any way to calculate the following exercise?

Question

I need to solve a problem and I have the following multiplication: $\frac {3*5*7*...*2015}{2*4*6*...*2014}-1$. Is there any way to solve it or I need to let it like this?

Answer 1

Multiplying numerator and denominator by $${2*4*6*...*2014}$$ gives $$\frac{2015!}{2^{2014}\cdot(1007!)^2}-1$$ Solving this isn't easy but Wolfram gives value $\approx 35.8204856288911755012-1=34.8204856288911755012 $

Answer 2

Generalizing, and with absolutely no originality,

$\begin{array}\\ \dfrac{3*5*...*(2n+1)}{2*4*...*2n} &=\dfrac{\prod_{k=1}^n (2k+1) }{\prod_{k=1}^n (2k)}\\ &=\dfrac{\prod_{k=1}^n (2k+1) }{\prod_{k=1}^n (2k)}\dfrac{\prod_{k=1}^n (2k)}{\prod_{k=1}^n (2k)}\\ &=\dfrac{\prod_{k=1}^{2n+1} k}{(\prod_{k=1}^n (2k))^2}\\ &=\dfrac{(2n+1)!}{4^n(n!)^2}\\ &\approx\dfrac{(2n+1)\sqrt{2\pi (2n)}(\frac{2n}{e})^{2n}(1+1/(24n))}{4^n(\sqrt{2\pi n}(\frac{n}{e})^n(1+1/(12n)))^2} \qquad\text{by Stirling}\\ &=\dfrac{(2n+1)\sqrt{4\pi n}}{4^n2\pi n}\dfrac{(2n)^{2n}}{e^{2n}}\dfrac{e^{2n}}{n^{2n}}\dfrac{1+1/(24n)}{1+1/(6n)+1/(144n^2)}\\ &\approx\dfrac{(2n+1)2\sqrt{\pi n}}{2\pi n}(1+\frac1{24n})(1-\frac1{6n})\\ &=\dfrac{(2n+1)}{\sqrt{\pi n}}(1-\frac1{8n})\\ \end{array} $

For $n=1007$ this is 35.82048535271531+ which differs by about 0.0000003.