+6248 \(\text{I have a simpler answer.}\\ \dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}+\dfrac{1}{(2k)^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}}=1\)
\(\sum \limits_{k=1}^\infty \dfrac{1}{(2k)^3} = \sum \limits_{k=1}^\infty \dfrac{1}{8k^3} = \dfrac 1 8 \sum \limits_{k=1}^\infty \dfrac{1}{k^3}\)
\(\dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}+\dfrac 1 8 \sum \limits_{k=1}^\infty\dfrac{1}{k^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}}=1\\ \dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}} + \dfrac 1 8 = 1\\ \dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}} = \dfrac 7 8\)
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Can anyone solve this?
