Math Problem Statement

mengapa bisa konvergen?

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

Solution

The series presented in the image is:

k=11(k+2)(k+3)\sum_{k=1}^{\infty} \frac{1}{(k+2)(k+3)}

Why it converges:

The proof uses partial fraction decomposition to express the terms as:

1(k+2)(k+3)=13(1k+21k+3)\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:

k=1n1(k+2)(k+3)=(131n+3)\sum_{k=1}^{n} \frac{1}{(k+2)(k+3)} = \left( \frac{1}{3} - \frac{1}{n+3} \right)

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

limn(131n+3)=13\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 13\frac{1}{3}, proving the series is convergent.

Would you like more detailed steps or have further questions?


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!

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Partial Fraction Decomposition
Telescoping Series

Formulas

\( \sum_{k=1}^{\infty} \frac{1}{(k+2)(k+3)} \)
Partial fraction: \( \frac{1}{(k+2)(k+3)} = \frac{1}{3} \left( \frac{1}{k+2} - \frac{1}{k+3} \right) \)

Theorems

Telescoping Series

Suitable Grade Level

Grades 11-12 or early university level