mysterious sum of two sequences Let 
$$S_1 = \sum_{n=1}^\infty \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \cdots$$
$$S_2 = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \cdots$$
So
\begin{align}
S_1 - S_2 = {} & 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \cdots \\
& {} - \left[ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \cdots \right] \\
= {} & 0 + 1 + 0 + \frac{1}{2} + 0 + \frac{1}{3} + 0 + \frac{1}{4} + \cdots \\
= {} & S_1
\end{align}
That makes $S_2$ zero, right?
The reason I am asking is because I was under the impression that $S_2 = \ln 2$. 
What did I miss?
Is it that $S_1$ is divergent and $S_2$ is convergent and it's not allowed?

You're saying $\infty - S_2 = \infty$; therefore $S_2=0$.
That is not valid.  If $f$ approaches $\infty$ then the limits of $f$ and $f-5$ are both the same, but that doesn't mean $5=0$.
