Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k}  = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series 
$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}  = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$
Does it converge?   If so, what is its sum?

it is not absolutely convergent (that is, if you are allowed to reorder terms you may end up with whatever number you fancy).
If you consider the associated series formed by summing the terms from 1 to n of the original one, that is you fix the order of summation of the original series, that series (which is not the original one...) converges to $\ln(2)$ See Wikipedia.
