Infinite product for sin z

Question

Assuming that $$sin z = {z}\prod_{r=1}^{\inf}(1-({z^2}/{r^2}{\pi^2})),$$ show that if $$m\rightarrow \inf$$ and $$ n\rightarrow \inf $$ in such a way that lim (m/n) = k where k is finite,then $$ lim\prod_{r=-n}^{m}’(1+(z/r\pi))={k^{z/\pi}}{sinz/z},$$the prime indicating that the factor for which r=0 is omitted.

Answer 1

The proof goes as follows.Multiply each term by $ e^{ z/r\pi}$ to get absolutely convergent terms. These can be rearranged into the sinz/z infinite product. The corresponding exponential terms have exponents(after combining the indices) $sum_{r=1}^m {z/r\pi}$,and $sum_{s=1}^n {-z/s\pi}$ Now let m,n go to infinity,the exponential sums go to log m plus the EM constant,and minus log n minus the EM constant,giving an exponent of ${(z/\pi)}(log(m/n) $= $log ({k}^{z/\pi}) $which gives the factor ${k}^{z/\pi }$in the answer.