We would like to find

\[\sum \limits_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(n+1)}\]

We use the generating series

\[F(z) = \sum \limits_{n=1}^{\infty} \frac{z^{n+1}}{n(n+1)}\]

We then take two derivatives to remove the $n(n+1)$, finding that

\[F''(z) = \sum \limits_{n=1}^{\infty} z^{n-1} = 1 + z + z^2 \dots\]

Then, using the geometric series formula

\[F''(z) = \frac{1}{1-z}\]

We see that $F (0) = 0$ and its well-known that $F(1) = \sum \limits_{n=1}^{\infty} \frac{1}{n(n+1)} = 1$. Using these boundary conditions, we can easily compute two integrals using some pretty standard methods

\[F(z) = z + (1-z)\ln(1-z)\]

We then see that

\[\boxed{F(-1) = 2 \ln (2) - 1}\]

Notice here how we’ve turned the computation of a sum into the solution of a differential equation with certain boundary conditions. We’ve basically turned a problem from one area of math (series) into another, albeit very closely related, area of math (differential equations).