Question:

Step 1: Identify the value of $a$. We recognize that $a$ is the Basel problem, a famous infinite series. The sum of reciprocals of perfect squares is: $$a = \sum_{r=1}^{\infty} \frac{1}{r^2} = \frac{\pi^2}{6}$$ Step 2: Identify the value of $b$. The series $b$ consists of reciprocals of squares of odd numbers: $$b = \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots = \frac{\pi^2}{8}$$ Step 3: Establish a relationship between $a$ and $b$. We can decompose the series $a$ into two parts: one containing odd-indexed terms and one containing even-indexed terms. $$a = \sum_{r=1}^{\infty} \frac{1}{r^2} = \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2} + \sum_{r=1}^{\infty} \frac{1}{(2r)^2}$$ This gives us: $$a = b + \sum_{r=1}^{\infty} \frac{1}{(2r)^2}$$ Step 4: Simplify the even-indexed series. The sum of reciprocals of squares of even numbers can be factored: $$\sum_{r=1}^{\infty} \frac{1}{(2r)^2} = \sum_{r=1}^{\infty} \frac{1}{4r^2} = \frac{1}{4}\sum_{r=1}^{\infty} \frac{1}{r^2} = \frac{1}{4}a$$ Step 5: Solve for the relationship between $a$ and $b$. Substituting back into the equation from Step 3: $$a = b + \frac{1}{4}a$$ Rearranging: $$a - \frac{1}{4}a = b$$ $$\frac{3}{4}a = b$$ Step 6: Calculate $\frac{3a}{b}$. From the relationship $\frac{3}{4}a = b$, we can write: $$\frac{3a}{b} = \frac{3a}{\frac{3}{4}a} = \frac{3a \cdot 4}{3a} = 4$$ Therefore, the value of $\frac{3a}{b} = \boxed{4}$, which corresponds to **Option 3**.