Question

If $ \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + ... \infty = \frac{\pi^4}{90}, $ $ \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + ... \infty = \alpha, $ $ \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + ... \infty = \beta, $ then $ \frac{\alpha}{\beta} $ is equal to:

• A. 23
• B. 15
• C. 14
• D. 18

Hint:

When dealing with sums of odd and even terms in infinite series, it can often be helpful to break the series into known components and apply symmetry or known formulas. In this case, the known formula for the sum of the reciprocals of powers was used to compute the result.

Answer 1

The Correct Option is B

To address this problem, we must examine the provided infinite series and their interrelationships:

The provided data includes:

• The series \(S = \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \ldots \infty\), which equals \( \frac{\pi^4}{90} \).
• The series comprising only odd terms, \( \alpha = \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \ldots \infty \).
• The series comprising only even terms, \( \beta = \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \ldots \infty \).

Our objective is to determine the ratio \( \frac{\alpha}{\beta} \).

Detailed Solution:

1. The complete series \( S \) can be segmented into the sum of the odd-term series and the even-term series:
   • \( S = \alpha + \beta \)
   • Given \( S = \frac{\pi^4}{90} \), it follows that \( \alpha + \beta = \frac{\pi^4}{90} \).
2. Let us analyze the even-term series \( \beta \):
   • This series can be written as: \( \beta = \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \ldots \)
   • By factoring out \( \frac{1}{2^4} \), we obtain:
     \( \beta = \frac{1}{2^4} \left(1 + \frac{1}{2^4} + \frac{1}{3^4} + \ldots\right) \)
   • The series within the parentheses shares the same structure as the original series \( S \), starting with \( \frac{1}{1^4} \).
   • Consequently, \( \beta = \frac{1}{16} \cdot S = \frac{1}{16} \cdot \frac{\pi^4}{90} = \frac{\pi^4}{1440} \).
3. Using the previously calculated values, we compute \( \alpha \):
   • \( \alpha = \frac{\pi^4}{90} - \beta = \frac{\pi^4}{90} - \frac{\pi^4}{1440} \)
   • To combine these fractions, we establish a common denominator:
     \( \alpha = \frac{\pi^4 \times 16}{1440} - \frac{\pi^4}{1440} = \frac{15\pi^4}{1440} \)
4. Finally, we calculate the ratio \( \frac{\alpha}{\beta} \):
   • \( \frac{\alpha}{\beta} = \frac{\frac{15\pi^4}{1440}}{\frac{\pi^4}{1440}} = 15 \)

Therefore, the ratio \( \frac{\alpha}{\beta} \) is determined to be 15.