To evaluate the sum of the infinite series:

$\sum_{n=3}^{\infty} \frac{1}{n^2 - 4}$

let’s break this down and analyze the series:

Step 1: Analyze the General Term

The general term is:

$\frac{1}{n^2 - 4} = \frac{1}{(n - 2)(n + 2)}$

This can be decomposed using partial fraction decomposition:

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

Step 2: Find the Partial Fractions

To find $A$ and $B$, we equate:

$1 = A(n + 2) + B(n - 2)$

Expanding and simplifying, we get:

$1 = A n + 2A + B n - 2B$

Combining like terms, this becomes:

$1 = (A + B)n + (2A - 2B)$

For the equality to hold for all $n$, we need:

1. $A + B = 0$
2. $2A - 2B = 1$

Step 3: Solve for $A$ and $B$

From $A + B = 0$, we have $B = -A$.

Substitute into the second equation:

$2A - 2(-A) = 1 \implies 4A = 1 \implies A = \frac{1}{4}$

Then, $B = -\frac{1}{4}$.

Step 4: Rewrite the Series

Using the values of $A$ and $B$, we get:

$\frac{1}{(n - 2)(n + 2)} = \frac{1/4}{n - 2} - \frac{1/4}{n + 2}$

So, the series can be written as:

$\sum_{n=3}^{\infty} \left( \frac{1}{4(n - 2)} - \frac{1}{4(n + 2)} \right)$

Step 5: Telescoping Series

This is a telescoping series, where most terms will cancel out. Let's expand the first few terms:

$\sum_{n=3}^{\infty} \left( \frac{1}{4(n - 2)} - \frac{1}{4(n + 2)} \right) = \frac{1}{4} \left( \frac{1}{1} - \frac{1}{5} + \frac{1}{2} - \frac{1}{6} + \frac{1}{3} - \frac{1}{7} + \cdots \right)$

Step 6: Sum of the Remaining Terms

In the telescoping series, all intermediate terms cancel, leaving us with the first few terms as the limit approaches infinity:

$\frac{1}{4} \left( 1 + \frac{1}{2} + \frac{1}{3} \right) = \frac{1}{4} \left( \frac{6}{6} + \frac{3}{6} + \frac{2}{6} \right) = \frac{1}{4} \cdot \frac{11}{6} = \frac{11}{24}$

Final Answer

The sum of the infinite series is:

$\sum_{n=3}^{\infty} \frac{1}{n^2 - 4} = \frac{11}{24}$

Would you like a more detailed explanation of any step, or do you have any other questions?

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Here are 5 related questions you might be interested in:

1. How do you perform partial fraction decomposition for more complex denominators?
2. What is a telescoping series and how can it simplify sums?
3. How do you evaluate the convergence of an infinite series?
4. What is the difference between convergent and divergent series?
5. How does the comparison test work for determining series convergence?

Tip: Telescoping series are often easier to sum because the terms cancel each other out, leaving only a few remaining terms to calculate.