What is the value of $$\lim_{n\to\infty}\prod_{k=1}^n\frac{k(k+1)+1+i}{k(k+1)+1-i}$$ where $i^2 = -1$? My guess is that it is equal to $i$.
My idea to prove it was to set $k = \tan\theta_k$ to obtain an expression with $e^{i\theta}$ and transform the product into a sum and get rid of as many terms as possible. I was not able to get down to a result.
Does someone have a solution? Thanks a lot in advance.
Easier way is: prove by induction on $n$ that $$ \prod_{k=1}^n\frac{k(k+1)+1+i}{k(k+1)+1-i} = i\;\frac{n+1-i}{n+1+i} $$