Question

Let $a = \sum_{r=1}^{\infty} \frac{1}{r^2}$ and $b = \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2}$. Then the value of $\frac{3a}{b}$ is equal to:

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

Answer: (C)

4

Answer 2

The given series are: $a = \sum_{r=1}^{\infty} \frac{1}{r^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots$ $b = \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2} = \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \dots$

We can express the series $a$ by separating terms with odd and even denominators: $a = \left(\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \dots\right) + \left(\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \dots\right)$

The first parenthesis is the series $b$. The second parenthesis is the sum of reciprocals of squares of even numbers. This can be written as: $\sum_{r=1}^{\infty} \frac{1}{(2r)^2} = \sum_{r=1}^{\infty} \frac{1}{4r^2} = \frac{1}{4} \sum_{r=1}^{\infty} \frac{1}{r^2} = \frac{1}{4}a$ Substituting these back into the expression for $a$: $a = b + \frac{1}{4}a$ Rearranging the equation to solve for $b$: $b = a - \frac{1}{4}a = \frac{3}{4}a$ We are asked to find the value of $\frac{3a}{b}$. Substitute the expression for $b$: $\frac{3a}{b} = \frac{3a}{\frac{3}{4}a}$ Since $a$ is a convergent series ($a = \frac{\pi^2}{6} \neq 0$), we can cancel $a$ from the numerator and denominator: $\frac{3a}{b} = \frac{3}{\frac{3}{4}} = 3 \times \frac{4}{3} = 4$