https://file.aiquickdraw.com/mathbot/file/1726486879781tck4qf23.jpg

The series presented in the image is:

\[
\sum_{k=1}^{\infty} \frac{1}{(k+2)(k+3)}
\]

### Why it converges:
The proof uses **partial fraction decomposition** to express the terms as:

\[
\frac{1}{(k+2)(k+3)} = \frac{1}{3} \left(\frac{1}{k+2} - \frac{1}{k+3}\right)
\]

This creates a **telescoping series**, where consecutive terms cancel out, leaving:

\[
\sum_{k=1}^{n} \frac{1}{(k+2)(k+3)} = \left( \frac{1}{3} - \frac{1}{n+3} \right)
\]

As \( n \to \infty \), the term \(\frac{1}{n+3}\) approaches 0. Therefore, the sum of the series approaches:

\[
\lim_{n \to \infty} \left( \frac{1}{3} - \frac{1}{n+3} \right) = \frac{1}{3}
\]

Thus, the series converges to a finite value, specifically \(\frac{1}{3}\), proving the series is convergent.

Would you like more detailed steps or have further questions?

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Here are some related questions to expand on this:

1. How does the partial fraction decomposition technique work in general?
2. Can all telescoping series be solved this way?
3. What are other examples of converging series?
4. How does the behavior of harmonic series compare to this?
5. What other tests can be used to determine the convergence of series?

**Tip**: Telescoping series are useful because they simplify complex terms by canceling out intermediate values!

Buktikan bahwa deret \( \sum_{k=1}^{\infty} \frac{1}{(k+2)(k+3)} \) konvergen!

Learn how to prove the convergence of the series \( \sum_{k=1}^{\infty} \frac{1}{(k+2)(k+3)} \) using partial fraction decomposition. This step-by-step solution includes key concepts like telescoping series and partial fractions. Suitable for advanced high school or early university students.

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