To solve the given infinite sum, we have:

$\sum_{n=5}^{\infty} \left( \frac{3n}{n+4} - \frac{3n - 3}{n+3} \right)$

Step-by-Step Solution

1. Separate Terms: Write the expression as two separate sums: $\sum_{n=5}^{\infty} \frac{3n}{n+4} - \sum_{n=5}^{\infty} \frac{3n - 3}{n+3}$

2. Simplify Each Fraction:

   • For $\frac{3n}{n+4}$, divide the numerator by the denominator: $\frac{3n}{n+4} = 3 - \frac{12}{n+4}$
   • For $\frac{3n - 3}{n+3}$, similarly divide: $\frac{3n - 3}{n+3} = 3 - \frac{9}{n+3}$

   This transforms our sum into: $\sum_{n=5}^{\infty} \left( 3 - \frac{12}{n+4} \right) - \sum_{n=5}^{\infty} \left( 3 - \frac{9}{n+3} \right)$

3. Combine the Expressions: Simplify by canceling out the constants $3$: $\sum_{n=5}^{\infty} \left( -\frac{12}{n+4} + \frac{9}{n+3} \right)$ This reduces to: $\sum_{n=5}^{\infty} \left( \frac{9}{n+3} - \frac{12}{n+4} \right)$

4. Recognize it as a Telescoping Series: In a telescoping series, many terms cancel out, leaving only a few terms from the beginning and end. Expanding a few terms will help see the pattern: $\left( \frac{9}{8} - \frac{12}{9} \right) + \left( \frac{9}{9} - \frac{12}{10} \right) + \left( \frac{9}{10} - \frac{12}{11} \right) + \dots$

   Most terms will cancel, and as $n \to \infty$, the remaining terms approach zero.

5. Calculate the Remaining Terms: Only the first terms, $\frac{9}{8}$ and $-\frac{12}{9}$, remain after cancellation: $\frac{9}{8} - \frac{12}{9}$

6. Simplify: Calculate the remaining expression: $\frac{9}{8} - \frac{12}{9} = \frac{81 - 96}{72} = -\frac{15}{72} = -\frac{5}{24}$

Final Answer

The value of the infinite sum is:

$\boxed{-\frac{5}{24}}$

Would you like further clarification on any of these steps?

Related Questions

1. What is a telescoping series, and how does it help in solving sums?
2. How do we know which terms will cancel in a telescoping series?
3. How can we simplify fractions like $\frac{3n}{n+4}$ in general?
4. What are other types of series that converge to a finite value?
5. How can we verify convergence of an infinite series?

Tip

In telescoping series, always expand the first few terms to identify cancellation patterns effectively.