sigma(infinity and n=5)(1/(x+1)-1/(n+2)) telescopic 

To evaluate the telescoping sum:

\[
\sum_{n=5}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+2} \right)
\]

This is a **telescoping series**, where terms cancel out as we expand the summation.

### Step-by-step solution:

1. **Write out the first few terms** of the sum:
   \[
   \left( \frac{1}{6} - \frac{1}{7} \right) + \left( \frac{1}{7} - \frac{1}{8} \right) + \left( \frac{1}{8} - \frac{1}{9} \right) + \dots
   \]

2. Notice the **cancellation** of intermediate terms:
   - The \(\frac{1}{7}\) in the first term cancels with \(-\frac{1}{7}\) in the second term.
   - The \(\frac{1}{8}\) in the second term cancels with \(-\frac{1}{8}\) in the third term, and so on.

3. As the sum progresses, all the terms cancel out except for the very first term of the series and the limiting term as \(n \to \infty\).

4. **At the limit as \(n \to \infty\):**
   \[
   \lim_{n \to \infty} \frac{1}{n+2} = 0
   \]
   
5. Therefore, the sum simplifies to:
   \[
   \sum_{n=5}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+2} \right) = \frac{1}{6}
   \]

### Final answer:
\[
\boxed{\frac{1}{6}}
\]

Would you like further details or have any questions on this? Here are 5 related questions to expand this topic:

1. How do you recognize a telescoping series in general?
2. What are other examples of telescoping series?
3. What happens when the series does not converge?
4. Can you apply telescoping sums to finite limits (non-infinite)?
5. What other types of series can be summed analytically?

**Tip:** Always check the behavior of the terms as \(n \to \infty\) to confirm whether a telescoping series converges or diverges.

sigma(infinity and n=5)(1/(x+1)-1/(n+2)) telescopic

Explore the step-by-step solution to a telescoping series problem where the sum of (1/(n+1) - 1/(n+2)) is calculated from n=5 to infinity. This advanced calculus problem includes limits and cancellation techniques. Suitable for college-level and advanced high school students.

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