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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.

Determine if the following series converges or diverges. If it converges, determine its sum: sum from n=2 to infinity of 1 / (n^2 - 1).

Explore how to determine the convergence of the series sum from n=2 to infinity of 1 / (n^2 - 1). This detailed solution includes partial fraction decomposition, a telescoping series approach, and the final sum calculation.

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