Given the series
$\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \cdots = \frac{\pi^4}{90},$
$\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \cdots = \alpha,$
$\frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \cdots = \beta.$ Then find $\frac{\alpha}{\beta}$
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When dealing with series involving only odd or even terms, try breaking the full sum into smaller parts based on the terms included and use known results like the Riemann zeta function.
Given the series \( \sum_{n=1}^\infty \frac{1}{n^4} = \zeta(4) = \frac{\pi^4}{90} \). - Let \( \alpha \) be the sum of the odd terms: \( \alpha = \sum_{k=1}^\infty \frac{1}{(2k-1)^4} \). Since \( \alpha \) comprises half the terms of the total series, \( \alpha = \frac{1}{2} \zeta(4) = \frac{1}{2} \cdot \frac{\pi^4}{90} = \frac{\pi^4}{180} \). - Let \( \beta \) be the sum of the even terms: \( \beta = \sum_{k=1}^\infty \frac{1}{(2k)^4} \). Factoring \( \frac{1}{2^4} = \frac{1}{16} \) from each term yields \( \beta = \frac{1}{16} \sum_{k=1}^\infty \frac{1}{k^4} = \frac{1}{16} \zeta(4) = \frac{1}{16} \cdot \frac{\pi^4}{90} = \frac{\pi^4}{1440} \). Calculating the ratio \( \frac{\alpha}{\beta} \): \[\frac{\alpha}{\beta} = \frac{\frac{\pi^4}{180}}{\frac{\pi^4}{1440}} = \frac{1440}{180} = 8.\]Therefore, the correct answer is 8.
