Find the value of $\dfrac{\cos(\pi/4)\cos(\pi/6)\cos(\pi/8)\cos(\pi/10)\cos(\pi/12)\cdots}{\cos(\pi/3)\cos(\pi/5)\cos(\pi/7)\cos(\pi/9)\cdots}$

Question

I manually calculated the values and found that the resulting answer was very close to $1.57\simeq \dfrac{\pi}{2}$

$$\dfrac{\cos(\pi/4)\cos(\pi/6)\cos(\pi/8)\cos(\pi/10)\cos(\pi/12)\cdots}{\cos(\pi/3)\cos(\pi/5)\cos(\pi/7)\cos(\pi/9)\cdots}=>\dfrac{\sin(\pi x/4)}{\sin(\pi x/6)}.\dfrac{\sin(\pi x/3)}{\sin( 3\pi x/10)}.\cdots$$

I assigned $x$ inside each term in order to separate and apply L' Hopital's rule for each term and somehow transform it into the Wallis product. But that may not be correct usage of the rule.

I do not know how to proceed further. I would appreciate any solutions either disproving or proving the assumed result of $\dfrac{\pi}{2}$

Answer 1

The answer is not $\pi/2$. It is $$A_\infty \approx 1.606992764721029112948.$$ Here's how to get so many digits of precision using a combination of numerical and computer algebra methods. Let $$ a_n = \frac{\cos\frac{\pi}{2n}}{\cos\frac{\pi}{2n-1}}\quad \text{and} \quad A_n = \prod_{k=2}^n a_k. $$ We can use computer algebra (in this case, Mathematica) to approximate $a_n$ for large $n$. We do this by substituting $n = 1/\epsilon$ into $a_n$, doing a series expansion, then substituting $\epsilon = 1/n$ into the result to find that $$ a_n = 1 + \frac{\pi^2}{8} n^{-3} + O(n^{-4}). $$ If there's a nice formula for the product of $1 + (\pi^2/8)k^{-3}$ from $k = n$ to $\infty$, then we could use this to approximate the part of the product we omit in a finite calculation. Mathematica does, in fact, provide an exact formula for this product. It's a little messy, so I won't write it out, but let's call it $R_n$. Then $$ A_\infty \approx A_n R_{n+1}. $$ Computing this for $n = 10^2$ to $n = 10^7$ (in $30$-digit-precision arithmetic), we observe that we gain three digits for precision each time we increase $n$ by a factor of $10$. This is why we get so many digits in the approximation stated above.

Finally, we can use the Inverse Symbolic Calculator to find out if this approximation of $A_\infty$ has some simple expression in terms of well-known constants. Apparently it does not.

Answer 2

Consider the partial product $$P_p=\frac{\prod _{n=2}^p \cos \left(\frac{\pi }{2 n}\right)}{\prod _{n=2}^p \cos \left(\frac{\pi }{2 n-1}\right)}=\frac{2 \sqrt{6}}{1+\sqrt{5}}\,\frac{\prod _{n=4}^p \cos \left(\frac{\pi }{2 n}\right)}{\prod _{n=4}^p \cos \left(\frac{\pi }{2 n-1}\right)}$$ Take logarithms, compose Taylor series from inside to outside $$\log \left(\cos \left(\frac{\pi }{p}\right)\right)=\sum_{m=2}^\infty (-1)^m\,\frac{2^{2 m-1} \left(4^m-1\right)\, \pi ^{2 m}\, B_{2 m}}{m \,(2 m)!}\,\,p^{-2m}$$ $$\log \left(\cos \left(\frac{\pi }{2 n}\right)\right)-\log \left(\cos \left(\frac{\pi }{2 n-1}\right)\right)=\frac {\pi^2}8 \sum_{k=3}^\infty \frac {a_k}{n^k}$$ where the first coefficients are

$$\left( \begin{array}{cc} n & a_n \\ 3 & 1 \\ 4 & \frac{3}{4} \\ 5 & \frac{6+\pi ^2}{12} \\ 6 & \frac{5(3+\pi ^2 )}{48} \\ 7 & \frac{45+25 \pi ^2+2 \pi ^4}{240} \\ 8 & \frac{7 \left(30+25 \pi ^2+4 \pi ^4\right)}{1920} \\ 9 & \frac{1260+1470 \pi ^2+392 \pi ^4+17 \pi ^6}{20160} \\ 10 & \frac{315+490 \pi ^2+196 \pi ^4+17 \pi ^6}{8960} \\ 11 & \frac{14175+28350 \pi ^2+15876 \pi ^4+2295 \pi ^6+62 \pi ^8}{725760} \\ 12 & \frac{11 \left(5670+14175 \pi ^2+10584 \pi ^4+2295 \pi ^6+124 \pi ^8\right)}{5806080} \\ \end{array} \right)$$ Using $$\sum_{n=4}^\infty \frac 1{n^k}=\zeta (k)-1-2^{-k}-3^{-k}$$ the summation write as a linear combination of zeta functions.

Using the terms given in the table $$P_\infty=\frac{2 \sqrt{6}}{1+\sqrt{5}} e^A\quad\text{where} \qquad A=0.05969631378607558355498878\cdots$$ that is to say $$P_\infty=1.60699218\cdots$$ which is identical to Wolfram Alpha value and very close to the exact number already given by @Jim Ferry.