To determine whether the series

$\sum_{n=1}^\infty \frac{-n!}{(n+2)!}$

converges or diverges, let us analyze it using series convergence tests and simplifying the general term.

Simplifying the General Term

The general term of the series is

$a_n = \frac{-n!}{(n+2)!}.$

Using the factorial property $(n+2)! = (n+2)(n+1)n!$, we can rewrite this as:

$a_n = \frac{-n!}{(n+2)(n+1)n!}.$

Canceling $n!$ from numerator and denominator, we get:

$a_n = \frac{-1}{(n+2)(n+1)}.$

So the series becomes:

$\sum_{n=1}^\infty \frac{-1}{(n+2)(n+1)}.$

Decomposing the General Term (Partial Fractions)

The term $\frac{-1}{(n+2)(n+1)}$ can be expressed using partial fractions:

$\frac{-1}{(n+2)(n+1)} = \frac{1}{n+1} - \frac{1}{n+2}.$

Thus, the series becomes a telescoping series:

$\sum_{n=1}^\infty \left(\frac{1}{n+1} - \frac{1}{n+2}\right).$

Telescoping Nature of the Series

Writing out the terms explicitly:

$\left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \cdots$

Most terms cancel, leaving:

$\sum_{n=1}^\infty \left(\frac{1}{n+1} - \frac{1}{n+2}\right) = \frac{1}{2} - \lim_{n \to \infty} \frac{1}{n+2}.$

As $n \to \infty$, $\frac{1}{n+2} \to 0$. Therefore, the sum converges to:

$\text{Sum} = \frac{1}{2}.$

Conclusion

The series $\sum_{n=1}^\infty \frac{-n!}{(n+2)!}$ converges.

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Would you like a deeper explanation of telescoping series or partial fractions?

Related Questions

1. What are other examples of telescoping series, and how are they solved?
2. How does the factorial grow compared to other sequences?
3. Can convergence tests like the ratio test apply to this series?
4. How does the harmonic series differ from telescoping series?
5. What is the significance of factorials in mathematical series?

Tip

For series involving factorials, simplifying using factorial properties and partial fractions often helps to reveal convergence behaviors.