If ` a = sum_(r=1)^(oo) 1/(r^(2)), b = sum_(r=1)^(oo) 1/((2r-1)^(2)),` then `(3a)/b` is equal to _____

Answer

To solve the problem, we need to find the value of \( \frac{3a}{b} \) where:

\[ a = \sum_{r=1}^{\infty} \frac{1}{r^2} \]
\[ b = \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2} \]

### Step 1: Evaluate \( a \)

The series \( a \) is known as the Basel problem, and its sum is given by:

\[ a = \sum_{r=1}^{\infty} \frac{1}{r^2} = \frac{\pi^2}{6} \]

### Step 2: Evaluate \( b \)

The series \( b \) consists of the reciprocals of the squares of the odd integers. We can express it as:

\[ b = \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2} \]

To find \( b \), we can relate it to \( a \). The sum of the squares of all integers can be split into the sum of the squares of odd integers and even integers:

\[ \sum_{r=1}^{\infty} \frac{1}{r^2} = \sum_{r=1}^{\infty} \frac{1}{(2r)^2} + \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2} \]

Calculating the sum of the squares of even integers:

\[ \sum_{r=1}^{\infty} \frac{1}{(2r)^2} = \sum_{r=1}^{\infty} \frac{1}{4r^2} = \frac{1}{4} \sum_{r=1}^{\infty} \frac{1}{r^2} = \frac{1}{4} \cdot \frac{\pi^2}{6} = \frac{\pi^2}{24} \]

Now substituting back into the equation:

\[ \frac{\pi^2}{6} = \frac{\pi^2}{24} + b \]

Solving for \( b \):

\[ b = \frac{\pi^2}{6} - \frac{\pi^2}{24} \]

Finding a common denominator (24):

\[ b = \frac{4\pi^2}{24} - \frac{\pi^2}{24} = \frac{3\pi^2}{24} = \frac{\pi^2}{8} \]

### Step 3: Calculate \( \frac{3a}{b} \)

Now we can substitute the values of \( a \) and \( b \):

\[ \frac{3a}{b} = \frac{3 \cdot \frac{\pi^2}{6}}{\frac{\pi^2}{8}} \]

Simplifying this expression:

\[ = \frac{3\pi^2}{6} \cdot \frac{8}{\pi^2} = \frac{3 \cdot 8}{6} = \frac{24}{6} = 4 \]

Thus, the final answer is:

\[ \frac{3a}{b} = 4 \]