`sum _(r = 2) ^(oo) (1)/(r ^(2) - 1)` is equal to :

Answer

To solve the series \( \sum_{r=2}^{\infty} \frac{1}{r^2 - 1} \), we will follow these steps:

### Step 1: Rewrite the Denominator
The term \( r^2 - 1 \) can be factored as:
\[ r^2 - 1 = (r - 1)(r + 1) \]
Thus, we can rewrite the series as:
\[ \sum_{r=2}^{\infty} \frac{1}{(r - 1)(r + 1)} \]

### Step 2: Use Partial Fraction Decomposition
Next, we can express \( \frac{1}{(r - 1)(r + 1)} \) using partial fractions:
\[ \frac{1}{(r - 1)(r + 1)} = \frac{A}{r - 1} + \frac{B}{r + 1} \]
Multiplying through by the denominator \( (r - 1)(r + 1) \) gives:
\[ 1 = A(r + 1) + B(r - 1) \]
Expanding this, we have:
\[ 1 = Ar + A + Br - B = (A + B)r + (A - B) \]
Setting up the equations:
1. \( A + B = 0 \)
2. \( A - B = 1 \)

### Step 3: Solve for A and B
From the first equation, \( B = -A \). Substituting into the second equation:
\[ A - (-A) = 1 \implies 2A = 1 \implies A = \frac{1}{2} \]
Thus, \( B = -\frac{1}{2} \). Therefore, we can rewrite:
\[ \frac{1}{(r - 1)(r + 1)} = \frac{1/2}{r - 1} - \frac{1/2}{r + 1} \]

### Step 4: Substitute Back into the Series
Now we substitute this back into our series:
\[ \sum_{r=2}^{\infty} \left( \frac{1/2}{r - 1} - \frac{1/2}{r + 1} \right) \]
This can be simplified as:
\[ \frac{1}{2} \sum_{r=2}^{\infty} \left( \frac{1}{r - 1} - \frac{1}{r + 1} \right) \]

### Step 5: Recognize the Telescoping Series
The series now becomes a telescoping series:
\[ \frac{1}{2} \left( \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \ldots \right) \]
Most terms will cancel out, and we are left with:
\[ \frac{1}{2} \left( 1 + \frac{1}{2} \right) \]

### Step 6: Calculate the Final Result
Calculating the remaining terms gives:
\[ \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4} \]

### Final Answer
Thus, the sum of the series is:
\[ \sum_{r=2}^{\infty} \frac{1}{r^2 - 1} = \frac{3}{4} \]