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Mathematics Question on Sequence and series

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

A

119\frac{11}{9}

B

11

C

−119\frac{-11}{9}

D

−113\frac{-11}{3}

Answer

−119\frac{-11}{9}

Explanation

Solution

The correct answer is (C) : −119\frac{-11}{9}
A = \sum_{n=1}^{\infty} $$\frac{1}{( 3 + (-1)^n)^n}
and
B = ∑n=1∞\sum_{n=1}^{\infty} (−1)n(3+(−1)n)n\frac{(-1)^n}{( 3 + (-1)^n)^n} ,
A = 12\frac{1}{2} ++ 142\frac{1}{4^2} ++ 123\frac{1}{2^3} ++ 144\frac{1}{4^4} ++ ......
B = −12\frac{-1}{2} ++ 142\frac{1}{4^2} −- 123\frac{1}{2^3} ++ 144\frac{1}{4^4} ++ ......
A = 121−14\frac{\frac{1}{2}}{1-\frac{1}{4}} ++ 1161−116\frac{\frac{1}{16}}{1-\frac{1}{16}} , B = −121−14\frac{-\frac{1}{2}}{1-\frac{1}{4}} ++ 1161−116\frac{\frac{1}{16}}{1-\frac{1}{16}}
A = 1115\frac{11}{15}, B = −915\frac{-9}{15}
∴ AB\frac{A}{B} = −119\frac{-11}{9}