If A = _n=1^(1)/((3+(-1)^n)^n) and B = _n=1^((-1)^n)/((3+(-1)^n)^n), then AB is equal to:

Question

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

Accepted Answer

$$A = \sum\limits_{n = 1}^\infty  {{1 \over {{{(3 + {{( - 1)}^n})}^n}}}} $$ and $$B = \sum\limits_{n = 1}^\infty  {{{{{( - 1)}^n}} \over {{{(3 + {{( - 1)}^n})}^n}}}} $$

$$A = {1 \over 2} + {1 \over {{4^2}}} + {1 \over {{2^3}}} + {1 \over {{4^4}}} + $$ ........

$$B = {{ - 1} \over 2} + {1 \over {{4^2}}} - {1 \over {{2^3}}} + {1 \over {{4^4}}} + $$ ......

$$A = {{{1 \over 2}} \over {1 - {1 \over 4}}} + {{{1 \over {16}}} \over {1 - {1 \over {16}}}}$$, $$B = {{ - {1 \over 2}} \over {1 - {1 \over 4}}} + {{{1 \over {16}}} \over {1 - {1 \over {16}}}}$$

$$A = {{11} \over {15}}$$, $$B = {{ - 9} \over {15}}$$

$$\therefore$$ $${A \over B} = {{ - 11} \over 9}$$

Solutions · 1