If `sum_(r=1)^(oo)(1)/((2r-1)^(2))=(pi^(2))/(8)`, then `sum_(r=1)^(oo) (1)/(r^(2))` is equal to

Options

A. `(pi^(2))/(24)`
B. `(pi^(2))/(3)`
C. `(pi^(2))/(6)`
D. none of these

Answer

To solve the problem, we start with the given information:

1. We know that:
\[ \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2} = \frac{\pi^2}{8} \]

2. We need to find:
\[ \sum_{r=1}^{\infty} \frac{1}{r^2} \]

### Step 1: Write the series for odd and even terms

The series \(\sum_{r=1}^{\infty} \frac{1}{r^2}\) can be split into two parts: the sum of the reciprocals of the squares of odd integers and the sum of the reciprocals of the squares of even integers.

- The sum of the squares of odd integers is given by:
\[ \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2} \]
- The sum of the squares of even integers is given by:
\[ \sum_{r=1}^{\infty} \frac{1}{(2r)^2} = \sum_{r=1}^{\infty} \frac{1}{4r^2} = \frac{1}{4} \sum_{r=1}^{\infty} \frac{1}{r^2} \]

### Step 2: Set up the equation

Now we can express the total sum of the series as:
\[ \sum_{r=1}^{\infty} \frac{1}{r^2} = \sum_{r=1}^{\infty} \frac{1}{(2r-1)^2} + \sum_{r=1}^{\infty} \frac{1}{(2r)^2} \]

Substituting the known values:
\[ \sum_{r=1}^{\infty} \frac{1}{r^2} = \frac{\pi^2}{8} + \frac{1}{4} \sum_{r=1}^{\infty} \frac{1}{r^2} \]

### Step 3: Let \( S = \sum_{r=1}^{\infty} \frac{1}{r^2} \)

Now we can rewrite the equation as:
\[ S = \frac{\pi^2}{8} + \frac{1}{4} S \]

### Step 4: Solve for \( S \)

To isolate \( S \), we can rearrange the equation:
\[ S - \frac{1}{4} S = \frac{\pi^2}{8} \]
\[ \frac{3}{4} S = \frac{\pi^2}{8} \]

Now, multiply both sides by \(\frac{4}{3}\):
\[ S = \frac{4}{3} \cdot \frac{\pi^2}{8} = \frac{\pi^2}{6} \]

### Conclusion

Thus, we find that:
\[ \sum_{r=1}^{\infty} \frac{1}{r^2} = \frac{\pi^2}{6} \]