Why does the 1/n term needs to be 0 in order to have absolute convergence for the product? $$P=\prod_{n=1}^{\infty }u_n=\prod \frac{(n-a_1)(n-a_2)...(n-a_n)}{(n-b_1)(n-b_2)...(n-b_n)}$$ $$u_n=1-\frac{a_1+a_2+...a_n-b_1-b_2-...b_n}{n}+O(n^{-2})$$
I also tried to solve it but I get $P=\coprod_{i=1}^{k}\frac{\Gamma (1-b_i)}{\Gamma (1-a_i)}e^\frac{b_i}{a_i} $ Could you show me the calculations in order to get the right answer? Whitch is $P=\coprod_{i=1}^{k}\frac{\Gamma (1-b_i)}{\Gamma (1-a_i)}$ and help calculate with this product: $pi /2=\prod_{n=1}^{\infty }\frac{4n^2}{4n^2-1}$
Tha infinite product $\prod (1+\alpha_n)$ converges absolutely (definition) iff the series $\sum \alpha_n$ converges absolutely.
In this case, the rational function $u_n$ satisfies $$ u_n = 1 + \frac{A}{n} + O\left(\frac{1}{n^2}\right)\qquad\text{as }n\to\infty $$ for some constant $A$. Now $u_n = 1+\alpha_n$. If $A \ne 0$, then $|\alpha_n| \sim \frac{|A|}{n}$, so $\sum |\alpha_n|$ diverges by comparison with the harmonic series $\sum \frac{1}{n}$. So $\sum \alpha_n$ does not converge absolutely. So $\prod u_n$ does not converge absolutely. So it remains only to compute $A$ in this case: $$ A = a_1 + a_2 + \dots+ a_k - b_1 - b_2 -\dots - b_k . $$ If this is not zero, then the product does not converge absolutely.