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

Answer

To solve the problem of finding the sum \( \sum_{r=2}^{\infty} \frac{1}{r^2 - 1} \), we can follow these steps:

### Step 1: Rewrite the term
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: Partial Fraction Decomposition
Next, we will use partial fraction decomposition to express \( \frac{1}{(r - 1)(r + 1)} \):
\[ \frac{1}{(r - 1)(r + 1)} = \frac{A}{r - 1} + \frac{B}{r + 1} \]
Multiplying through by \( (r - 1)(r + 1) \) gives:
\[ 1 = A(r + 1) + B(r - 1) \]
Expanding this, we get:
\[ 1 = Ar + A + Br - B = (A + B)r + (A - B) \]
Setting the coefficients equal, we have:
1. \( A + B = 0 \)
2. \( A - B = 1 \)

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 write:
\[ \frac{1}{(r - 1)(r + 1)} = \frac{1/2}{r - 1} - \frac{1/2}{r + 1} \]

### Step 3: Rewrite the series
Now substituting back into the series:
\[ \sum_{r=2}^{\infty} \left( \frac{1/2}{r - 1} - \frac{1/2}{r + 1} \right) \]
This can be factored out:
\[ \frac{1}{2} \sum_{r=2}^{\infty} \left( \frac{1}{r - 1} - \frac{1}{r + 1} \right) \]

### Step 4: Recognize the telescoping series
The series is telescoping:
\[ \frac{1}{2} \left( \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \ldots \right) \]
Most terms will cancel out, and we are left with:
\[ \frac{1}{2} \left( 1 + \frac{1}{2} - \lim_{n \to \infty} \frac{1}{n + 1} - \lim_{n \to \infty} \frac{1}{n + 2} \right) \]
As \( n \to \infty \), both limits approach 0, thus we have:
\[ \frac{1}{2} \left( 1 + \frac{1}{2} \right) = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4} \]

### Final Answer
Therefore, the sum \( \sum_{r=2}^{\infty} \frac{1}{r^2 - 1} \) is equal to:
\[ \frac{3}{4} \]