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The image shows a mathematical expression that needs to be proven. The equation appears to be:

\[
\sum_{r=1}^{n} \frac{1}{(r^2 - 1)} = \frac{n}{2n+1}
\]

This is a summation that involves terms of the form \( \frac{1}{r^2 - 1} \), and the goal is to prove that the sum from \( r = 1 \) to \( n \) is equal to the right-hand side expression \( \frac{n}{2n+1} \).

### Steps to prove this:
1. First, recognize that \( \frac{1}{r^2 - 1} \) can be factored using the difference of squares:
   \[
   \frac{1}{r^2 - 1} = \frac{1}{(r-1)(r+1)}
   \]

2. This expression can be decomposed into partial fractions:
   \[
   \frac{1}{(r-1)(r+1)} = \frac{1}{2} \left( \frac{1}{r-1} - \frac{1}{r+1} \right)
   \]

3. Therefore, the original sum becomes:
   \[
   \sum_{r=1}^{n} \frac{1}{(r^2 - 1)} = \frac{1}{2} \sum_{r=1}^{n} \left( \frac{1}{r-1} - \frac{1}{r+1} \right)
   \]

4. This is now a telescoping series, where many terms cancel out, leaving only a few terms from the beginning and end of the summation.

5. After simplifying, you will get the final expression for the sum, which is \( \frac{n}{2n+1} \), thus proving the equation.

Would you like further details on any specific steps or a deeper breakdown of the proof?

Here are 5 follow-up questions:
1. What is a telescoping series, and how does it simplify complex summations?
2. How are partial fractions useful in simplifying sums or integrals?
3. What other types of summations can be simplified using the difference of squares?
4. Can you derive a similar formula for different types of quadratic denominators?
5. How does the convergence of such series behave as \( n \) increases?

**Tip**: Partial fractions are a powerful technique in simplifying sums involving rational functions, especially when the terms can telescope.

Prove that the sum of the series \( \sum_{r=1}^{n} \frac{1}{r^2 - 1} = \frac{n}{2n + 1} \).

This proof shows how to evaluate the summation \( \sum_{r=1}^{n} \frac{1}{r^2 - 1} \) using partial fractions and telescoping series, resulting in the expression \( \frac{n}{2n+1} \). Detailed step-by-step breakdown included.

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