Example 3.53 in Baby Rudin
We can show that the series converges, and find its sum, as follows:
$\hspace{.3 in}1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots=\ln 2$ $\;\;\;$so
$\hspace{.27 in}\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots=\frac{1}{2}\ln 2$. $\;\;\;$Inserting zeros, we get
$\hspace{.26 in}0+\frac{1}{2}+0-\frac{1}{4}+0+\frac{1}{6}+0-\frac{1}{8}+0+\frac{1}{10}+\cdots=\frac{1}{2}\ln 2$.
Adding this to the original series gives
$\hspace{.26 in}1+0+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+0+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+0+\cdots=\frac{3}{2}\ln 2$, $\;\;$ so
$\hspace{.25 in} 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\cdots=\frac{3}{2}\ln 2$.