We are asked to show that the series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n(n+1)}$ and $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{2^n}{n!}$ converge absolutely. A series $\sum_{n=1}^{\infty} a_n$ converges absolutely if $\sum_{n=1}^{\infty} |a_n|$ converges.

1. Problem Description

We are asked to show that the series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n(n+1)}$ and $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{2^n}{n!}$ converge absolutely. A series $\sum_{n=1}^{\infty} a_n$ converges absolutely if $\sum_{n=1}^{\infty} |a_n|$ converges.

2. Solution Steps

For the first series, $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n(n+1)}$, we need to examine the series $\sum_{n=1}^{\infty} \left| (-1)^{n+1} \frac{1}{n(n+1)} \right| = \sum_{n=1}^{\infty} \frac{1}{n(n+1)}$.

We can use the limit comparison test. Let $a_n = \frac{1}{n(n+1)}$ and $b_n = \frac{1}{n^2}$.

Then, $\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{1}{n(n+1)}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^2}{n(n+1)} = \lim_{n \to \infty} \frac{n^2}{n^2+n} = \lim_{n \to \infty} \frac{1}{1+\frac{1}{n}} = 1$.

Since the limit is a finite positive number, and we know that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges (p-series with $p=2>1$), then $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ also converges.

Therefore, the series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n(n+1)}$ converges absolutely.

For the second series, $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{2^n}{n!}$, we need to examine the series $\sum_{n=1}^{\infty} \left| (-1)^{n+1} \frac{2^n}{n!} \right| = \sum_{n=1}^{\infty} \frac{2^n}{n!}$.

We can use the ratio test. Let $a_n = \frac{2^n}{n!}$.

Then, $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}} = \lim_{n \to \infty} \frac{2^{n+1} n!}{2^n (n+1)!} = \lim_{n \to \infty} \frac{2^{n} \cdot 2 \cdot n!}{2^n \cdot (n+1) n!} = \lim_{n \to \infty} \frac{2}{n+1} = 0$.

Since the limit is less than 1, the series $\sum_{n=1}^{\infty} \frac{2^n}{n!}$ converges.

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

3. Final Answer

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

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