If $$A = \sum_{n=1}^{\infty} \frac{1}{(3+(- 1)^n)^n}$$ and $$B = \sum_{n=1}^{\infty} \frac{(-1)^n}{(3+(-1)^n)^n}$$, then $$\frac{A}{B}$$ is equal to

We need to find $$\frac{A}{B}$$ where $$A = \sum_{n=1}^{\infty} \frac{1}{(3+(-1)^n)^n}$$ and $$B = \sum_{n=1}^{\infty} \frac{(-1)^n}{(3+(-1)^n)^n}$$.

When n is odd: $$(-1)^n = -1$$, so $$(3+(-1)^n)^n = 2^n$$.

When n is even: $$(-1)^n = 1$$, so $$(3+(-1)^n)^n = 4^n$$.

$$A = \sum_{\text{odd } n} \frac{1}{2^n} + \sum_{\text{even } n} \frac{1}{4^n}$$

$$A = \left(\frac{1}{2} + \frac{1}{2^3} + \frac{1}{2^5} + \cdots\right) + \left(\frac{1}{4^2} + \frac{1}{4^4} + \frac{1}{4^6} + \cdots\right)$$

$$A = \frac{1/2}{1 - 1/4} + \frac{1/16}{1 - 1/16} = \frac{1/2}{3/4} + \frac{1/16}{15/16} = \frac{2}{3} + \frac{1}{15} = \frac{10 + 1}{15} = \frac{11}{15}$$

For odd n: $$(-1)^n = -1$$. For even n: $$(-1)^n = 1$$.

$$B = \sum_{\text{odd } n} \frac{-1}{2^n} + \sum_{\text{even } n} \frac{1}{4^n} = -\frac{2}{3} + \frac{1}{15} = \frac{-10 + 1}{15} = -\frac{9}{15} = -\frac{3}{5}$$

$$\frac{A}{B} = \frac{11/15}{-3/5} = \frac{11}{15} \times \frac{-5}{3} = -\frac{55}{45} = -\frac{11}{9}$$

The answer is Option C: $$-\dfrac{11}{9}$$.