<s>[INST] 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?
 [/INST] You can't rearrange the terms of a sequence that doesn't converge absolutely without changing it's value, and you can't group series like that. For example:
$$
\sum_{i=0}^\infty (-1)^i = 1 -1 + 1 - 1 +\cdots \ne (1-1) + (1-1) + \cdots 
$$
which would make it look like the sum converges to zero, but the above sum clearly diverges.
</s>