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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.

Determine the value of the following infinite sum: \( \sum_{n=5}^{\infty} \left( \frac{3n}{n+4} - \frac{3n-3}{n+3} \right) \).

Solve the infinite series \( \sum_{n=5}^{\infty} \left( \frac{3n}{n+4} - \frac{3n-3}{n+3} \right) \) by simplifying using algebraic techniques and identifying the telescoping pattern. Learn how terms cancel and obtain the finite result for this advanced math problem.

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