If A = ∑ n = 1 ∞ 1 3 + - 1 n n and B = ∑ n = 1 ∞ - 1 n 3 + - 1 n n , then A B is…

Given, $A=\sum _{n=1}^{\infty }\frac{1}{{\left(3+{\left(-1\right)}^n\right)}^n}=\frac{1}{2}+\frac{1}{4^2}+\frac{1}{2^3}+\frac{1}{4^4}+\cdots$

$=\left(\frac{1}{2}+\frac{1}{2^3}+\cdots \right)+\left(\frac{1}{4^2}+\frac{1}{4^4}+\cdots \right)$$=\frac{\frac{1}{2}}{1-\frac{1}{4}}+\frac{\frac{1}{16}}{1-\frac{1}{16}}$

$=\frac{2}{3}+\frac{1}{15}=\frac{11}{15}$

$B=\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n}{{\left(3+{\left(-1\right)}^n\right)}^n}=-\frac{1}{2}+\frac{1}{4^2}-\frac{1}{2^3}+\frac{1}{4^4}-\cdots$

$=-\left(\frac{1}{2}+\frac{1}{2^3}+\cdots \right)+\left(\frac{1}{4^2}+\frac{1}{4^4}+\cdots \right)$

$=-\frac{\frac{1}{2}}{1-\frac{1}{4}}+\frac{\frac{1}{16}}{1-\frac{1}{16}}=-\frac{2}{3}+\frac{1}{15}=-\frac{9}{15}$

Hence, $\frac{A}{B}=-\frac{11}{9}$