Question

Given that \( 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots = \frac{\pi^2}{6} \), the value of \( 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \dots \) is

• A. \( \frac{\pi^2}{6} - 1 \)
• B. \( \frac{\pi^2}{12} \)
• C. \( \frac{\pi^2}{8} \)
• D. \( \frac{\pi}{6} \)

Hint:

This is a standard result worth remembering. The sum of the reciprocals of the squares of all natural numbers is \( \pi^2/6 \). The sum for just the odd numbers is \( \pi^2/8 \), and the sum for just the even numbers is \( \pi^2/24 \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem involves manipulating the famous Basel problem series sum. The given series includes all natural numbers. The series we need to find includes only the odd natural numbers. The key is to separate the given series into its odd and even components.
Step 2: Key Formula or Approach:
Let \( S = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \). We can split S into the sum of odd terms and even terms: \[ S = \left( \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \dots \right) + \left( \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \dots \right) \] Let \( S_{odd} \) be the sum of the odd terms (which we want to find) and \( S_{even} \) be the sum of the even terms. \[ S = S_{odd} + S_{even} \] We can express \( S_{even} \) in terms of S.
Step 3: Detailed Explanation:
Let's analyze the sum of the even terms, \( S_{even} \): \[ S_{even} = \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \dots = \sum_{k=1}^{\infty} \frac{1}{(2k)^2} \] \[ S_{even} = \sum_{k=1}^{\infty} \frac{1}{4k^2} \] We can factor out the constant \( \frac{1}{4} \) from the summation: \[ S_{even} = \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{k^2} \] The summation \( \sum_{k=1}^{\infty} \frac{1}{k^2} \) is just the original series S. So, \( S_{even} = \frac{1}{4} S \). Now substitute this back into the split equation: \[ S = S_{odd} + S_{even} \] \[ S = S_{odd} + \frac{1}{4} S \] We can now solve for \( S_{odd} \): \[ S_{odd} = S - \frac{1}{4} S = \frac{3}{4} S \] We are given that \( S = \frac{\pi^2}{6} \). \[ S_{odd} = \frac{3}{4} \left( \frac{\pi^2}{6} \right) = \frac{3\pi^2}{24} = \frac{\pi^2}{8} \] Step 4: Final Answer:
The value of the series \( 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \dots \) is \( \frac{\pi^2}{8} \).