We are tasked to determine whether the telescoping series
$\sum_{n=2}^\infty \frac{6}{n^2 - 1}$ is convergent or divergent, and if convergent, find its sum.

Step 1: Factorize the denominator

We start by factorizing $n^2 - 1$ as a difference of squares:
$n^2 - 1 = (n-1)(n+1).$ Thus, the general term becomes:
$\frac{6}{n^2 - 1} = \frac{6}{(n-1)(n+1)}.$

Step 2: Decompose using partial fractions

We write:
$\frac{6}{(n-1)(n+1)} = \frac{A}{n-1} + \frac{B}{n+1},$ where $A$ and $B$ are constants to be determined.

Multiplying through by $(n-1)(n+1)$, we get:
$6 = A(n+1) + B(n-1).$ Expanding and collecting terms:
$6 = An + A + Bn - B \quad \implies \quad 6 = (A+B)n + (A-B).$ By comparing coefficients of $n$ and the constant term:

• $A + B = 0$ (coefficient of $n$),
• $A - B = 6$ (constant term).

From $A + B = 0$, we get $B = -A$. Substituting into $A - B = 6$:
$A - (-A) = 6 \quad \implies \quad 2A = 6 \quad \implies \quad A = 3.$ Thus, $B = -3$.

So the partial fraction decomposition is:
$\frac{6}{(n-1)(n+1)} = \frac{3}{n-1} - \frac{3}{n+1}.$

Step 3: Write the series in telescoping form

Substituting into the series:
$\sum_{n=2}^\infty \frac{6}{n^2 - 1} = \sum_{n=2}^\infty \left( \frac{3}{n-1} - \frac{3}{n+1} \right).$

Expanding a few terms:
$\left( \frac{3}{1} - \frac{3}{3} \right) + \left( \frac{3}{2} - \frac{3}{4} \right) + \left( \frac{3}{3} - \frac{3}{5} \right) + \dots$ Notice that most terms cancel out, leaving only the first few terms of the series.

Step 4: Simplify the remaining terms

The cancellation leaves:
$3 \left( 1 + \frac{1}{2} \right).$

Thus, the sum of the series is:
$3 \left( 1 + \frac{1}{2} \right) = 3 \cdot \frac{3}{2} = \frac{9}{2}.$

Step 5: Conclusion

The series converges and its sum is:
$\boxed{\frac{9}{2}}$

Would you like more explanation or another related example?

Here are some related questions for practice:

1. Prove the convergence of $\sum_{n=2}^\infty \frac{1}{n^2}$.
2. Evaluate the sum of $\sum_{n=1}^\infty \frac{n}{n^2 + 1}$.
3. Use partial fractions to decompose $\frac{1}{n^3 - n}$.
4. Show that the harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ diverges.
5. Find the sum of $\sum_{n=1}^\infty \frac{6}{n^2 + 3n + 2}$.

Tip: Telescoping series simplify by partial fraction decomposition, leading to cancellations in consecutive terms. Always identify patterns in the remaining terms!