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?

$\sum_{k=1}^{n} ( \frac{1}{2k-1}-\frac{1}{2k} ) = \sum_{k=1}^{n} ( \frac{1}{2k-1}+\frac{1}{2k} ) - 2 \sum_{k=1}^{n} \frac{1}{2k} = \sum_{k=1}^{2n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k} = \sum_{k=n+1}^{2n} \frac{1}{k}$.
$\ln(2) \overset{n\to\infty}{\leftarrow} \ln(2) + \ln(\frac{2n+1}{2n+2}) = \ln(2n+1)-\ln(n+1)$
$= \int_{n+1}^{2n+1} \frac{1}{x}\ dx \le \sum_{k=n+1}^{2n} \frac{1}{k} \le \int_{n}^{2n} \frac{1}{x}\ dx$
$= \ln(2n)-\ln(n) = \ln(2)$.
So by squeeze theorem we are done.
