Key Concepts and Formulas

• Decomposition of Series: An infinite series can be split into the sum of its terms with odd indices and its terms with even indices: $S = \sum_{n=1}^\infty a_n = \sum_{k=1}^\infty a_{2k-1} + \sum_{k=1}^\infty a_{2k}$.
• Exponent Properties: $(ab)^m = a^m b^m$.
• Algebraic Manipulation: Factoring and simplifying expressions involving series.

Step-by-Step Solution

We are given the following series:

1. $S = \sum_{n=1}^\infty \frac{1}{n^4} = \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \ldots = \frac{\pi^4}{90}$
2. $\alpha = \sum_{k=1}^\infty \frac{1}{(2k-1)^4} = \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \ldots$
3. $\beta = \sum_{k=1}^\infty \frac{1}{(2k)^4} = \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \ldots$

Our goal is to find the ratio $\frac{\alpha}{\beta}$.

Step 1: Relate $S$, $\alpha$, and $\beta$ using the decomposition of series. The series $S$ is the sum of all terms $\frac{1}{n^4}$ for $n=1, 2, 3, \ldots$. This can be split into the sum of terms where $n$ is odd and the sum of terms where $n$ is even. The sum of terms where $n$ is odd is precisely $\alpha$. The sum of terms where $n$ is even is precisely $\beta$. Therefore, we can write: $S = \alpha + \beta$

Step 2: Simplify the series $\beta$ and express it in terms of $S$. The series $\beta$ consists of terms with even denominators: $\beta = \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \ldots$ Each term in the denominator is of the form $(2k)^4$. We can rewrite $\beta$ as: $\beta = \sum_{k=1}^\infty \frac{1}{(2k)^4}$ Using the exponent property $(ab)^m = a^m b^m$, we have $(2k)^4 = 2^4 k^4$. $\beta = \sum_{k=1}^\infty \frac{1}{2^4 k^4}$ We can factor out the constant term $\frac{1}{2^4}$ from the summation: $\beta = \frac{1}{2^4} \sum_{k=1}^\infty \frac{1}{k^4}$ The summation $\sum_{k=1}^\infty \frac{1}{k^4}$ is exactly the definition of $S$. So, we have: $\beta = \frac{1}{16} S$

Step 3: Express $\alpha$ in terms of $S$. From Step 1, we have the relationship $S = \alpha + \beta$. Substitute the expression for $\beta$ from Step 2 into this equation: $S = \alpha + \frac{1}{16} S$ Now, we solve for $\alpha$: $\alpha = S - \frac{1}{16} S$ $\alpha = \left(1 - \frac{1}{16}\right) S$ $\alpha = \frac{15}{16} S$

Step 4: Calculate the ratio $\frac{\alpha}{\beta}$. We have $\alpha = \frac{15}{16} S$ and $\beta = \frac{1}{16} S$. Now, we find the ratio: $\frac{\alpha}{\beta} = \frac{\frac{15}{16} S}{\frac{1}{16} S}$ Since $S = \frac{\pi^4}{90}$ is a non-zero constant, we can cancel $S$ from the numerator and denominator. The value of $S$ is not needed to find the ratio. $\frac{\alpha}{\beta} = \frac{\frac{15}{16}}{\frac{1}{16}}$ Multiply the numerator and the denominator by 16 to simplify: $\frac{\alpha}{\beta} = 15$

The value of $\frac{\alpha}{\beta}$ is 15.

Common Mistakes & Tips

• Confusing Series: Be careful to distinguish between the sum of all terms, the sum of odd-indexed terms, and the sum of even-indexed terms.
• Algebraic Errors: Ensure accuracy when factoring out common terms and simplifying fractions. A small mistake can lead to a completely wrong answer.
• Unnecessary Information: The specific value of $S = \frac{\pi^4}{90}$ is provided as context but is not required to solve for the ratio. Focus on the relationships between the series.

Summary

This problem is solved by recognizing that the sum of all terms in the series ($S$) can be decomposed into the sum of odd-indexed terms ($\alpha$) and even-indexed terms ($\beta$). By manipulating the series for $\beta$, we found it to be a simple fraction of $S$. This allowed us to express both $\alpha$ and $\beta$ in terms of $S$, leading to a straightforward calculation of their ratio.

The final answer is $\boxed{15}$.