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

Step 1: General Series Formula We are given the series: \[ S = \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}. \] This series is the standard series for the sum of the reciprocals of the 4th powers of natural numbers.
Step 2: Express \( \alpha \) and \( \beta \) We need to find the values of \( \alpha \) and \( \beta \), which are defined as follows: - \( \alpha = \sum_{n=1, \, n \text{ odd}}^{\infty} \frac{1}{n^4} \) - \( \beta = \sum_{n=1, \, n \text{ even}}^{\infty} \frac{1}{n^4} \) The total sum \( S \) can be split into the sum of the odd and even terms: \[ S = \alpha + \beta. \] From the problem statement, we know: \[ S = \frac{\pi^4}{90}. \]
Step 3: Breaking the Series Into Odd and Even Terms We can now express the sum \( \alpha \) and \( \beta \) in terms of the standard sum for the series of the 4th powers. For \( \alpha \) (odd terms): \[ \alpha = \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \dots. \] For \( \beta \) (even terms): \[ \beta = \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \dots. \]
Step 4: Known Results for Sums of Odd and Even Reciprocals It is known that the sum of the even terms can be related to the full series by factoring out the powers of 2: \[ \beta = \frac{1}{16} \left( \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \dots \right) = \frac{1}{16} \cdot \frac{\pi^4}{90}. \] Thus: \[ \beta = \frac{\pi^4}{1440}. \] Now, we can substitute this into the equation for the total sum \( S \): \[ S = \alpha + \beta = \frac{\pi^4}{90}. \] Therefore: \[ \alpha = \frac{\pi^4}{90} - \frac{\pi^4}{1440} = \frac{16\pi^4}{1440} - \frac{\pi^4}{1440} = \frac{15\pi^4}{1440} = \frac{\pi^4}{96}. \]
Step 5: Finding \( \frac{\alpha}{\beta} \) Now we can compute \( \frac{\alpha}{\beta} \): \[ \frac{\alpha}{\beta} = \frac{\frac{\pi^4}{96}}{\frac{\pi^4}{1440}} = \frac{1440}{96} = 15. \]