The problem asks whether the series

$\sum_{n=2}^{\infty} \frac{1}{n^2 - 1}$

converges or diverges. If it converges, we are asked to find its sum.

Step 1: Simplifying the general term

We begin by simplifying the general term $\frac{1}{n^2 - 1}$, which can be factored as:

$n^2 - 1 = (n - 1)(n + 1)$

Thus, the term becomes:

$\frac{1}{n^2 - 1} = \frac{1}{(n-1)(n+1)}$

This is a partial fraction decomposition. We can express this as:

$\frac{1}{(n-1)(n+1)} = \frac{A}{n-1} + \frac{B}{n+1}$

To find $A$ and $B$, we solve the equation:

$\frac{1}{(n-1)(n+1)} = \frac{A}{n-1} + \frac{B}{n+1}$

Multiplying through by $(n-1)(n+1)$ gives:

$1 = A(n+1) + B(n-1)$

Expanding this gives:

$1 = A n + A + B n - B$

This simplifies to:

$1 = (A + B)n + (A - B)$

Equating coefficients of $n$ and the constant terms, we get the system:

$A + B = 0 \quad \text{and} \quad A - B = 1$

Solving these equations gives $A = \frac{1}{2}$ and $B = -\frac{1}{2}$.

Thus, we can rewrite the general term as:

$\frac{1}{(n-1)(n+1)} = \frac{1}{2}\left(\frac{1}{n-1} - \frac{1}{n+1}\right)$

Step 2: Recognizing the telescoping series

The series now becomes:

$\sum_{n=2}^{\infty} \frac{1}{n^2 - 1} = \frac{1}{2} \sum_{n=2}^{\infty} \left( \frac{1}{n-1} - \frac{1}{n+1} \right)$

This is a telescoping series, meaning most terms will cancel out. Let's write out the first few terms to see this:

$\frac{1}{2} \left( \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \cdots \right)$

Notice that many terms cancel out, and we are left with:

$\frac{1}{2} \left( 1 + \frac{1}{2} \right)$

Step 3: Final sum

Thus, the remaining terms are:

$\frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$

Conclusion

The series converges, and its sum is $\frac{3}{4}$.

Would you like more details or have any questions?

Related Questions:

1. How can we use the comparison test to check if a series converges?
2. What is a telescoping series, and when does it occur?
3. What are other methods to check the convergence of a series?
4. How does partial fraction decomposition work in series?
5. Can all rational functions be decomposed into partial fractions?

Tip: Whenever you encounter a rational function in a series, always check if partial fractions can simplify the terms. This often leads to a telescoping series, which can simplify finding the sum.