We need to find the sum of the series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(n+1)}$.

1. Problem Description

We need to find the sum of the series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(n+1)}$.

2. Solution Steps

First, we decompose the fraction $\frac{1}{n(n+1)}$ using partial fractions.

$\frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$.

Multiplying both sides by $n(n+1)$, we get

$1 = A(n+1) + Bn$.

When $n = 0$, we have $1 = A(0+1) + B(0)$, so $A = 1$.

When $n = -1$, we have $1 = A(-1+1) + B(-1)$, so $-B = 1$, which means $B = -1$.

Therefore,

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

Now we can rewrite the series as

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

We know that the alternating harmonic series is $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln(2)$.

Now, consider the second summation:

$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1} = \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \cdots$

$= 1 - (1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots) = 1 - \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - \ln(2)$.

Then the original sum becomes:

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

2. Final Answer

$2\ln(2) - 1$