If $$A = \sum\limits_{n = 1}^\infty {{1 \over {{{\left( {3 + {{( - 1)}^n}} \right)}^n}}}} $$ and $$B = \sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{{\left( {3 + {{( - 1)}^n}} \right)}^n}}}} $$, then ${A \over B}$ is equal to :
Solution
<p>$A = \sum\limits_{n = 1}^\infty {{1 \over {{{(3 + {{( - 1)}^n})}^n}}}}$ and $$B = \sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{{(3 + {{( - 1)}^n})}^n}}}} $$</p>
<p>$A = {1 \over 2} + {1 \over {{4^2}}} + {1 \over {{2^3}}} + {1 \over {{4^4}}} +$ ........</p>
<p>$$B = {{ - 1} \over 2} + {1 \over {{4^2}}} - {1 \over {{2^3}}} + {1 \over {{4^4}}} + $$ ......</p>
<p>$$A = {{{1 \over 2}} \over {1 - {1 \over 4}}} + {{{1 \over {16}}} \over {1 - {1 \over {16}}}}$$, $$B = {{ - {1 \over 2}} \over {1 - {1 \over 4}}} + {{{1 \over {16}}} \over {1 - {1 \over {16}}}}$$</p>
<p>$A = {{11} \over {15}}$, $B = {{ - 9} \over {15}}$</p>
<p>$\therefore$ ${A \over B} = {{ - 11} \over 9}$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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