Show that $\sum (-1)^{n+1}/n$ = $k$ can be expanded to a sum with $ 1 - 1/2 + 1/3 + 1/5 - 1/4 +1/7 + 1/9 - 1/6 + ....$ which is $3k/2$ I need to show that $\sum (-1)^{n+1} / n$ = $k$ (which is $1 - 1/2 + 1/3 - 1/4+...$) can be expanded to a sum with $ 1 - 1/2 + 1/3 + 1/5 - 1/4 +1/7 + 1/9 - 1/6 + .... $ which is $3k/2$. 
How can I start? I know that $k/2$ = $\sum (-1)^{n+1}/2n$ How can I go further so I get $3k/2$?
 A: Are you asking whether
$$\sum_{k=1}^\infty\left(\frac1{4k-3}-\frac1{2k}+\frac1{4k-1}\right)=
\frac32\ln2?$$
Observe that
$$\sum_{k=1}^n\left(\frac1{4k-3}-\frac1{2k}+\frac1{4k-1}\right)
=H_{4n}-\frac{H_{2n}}{2}-\frac{H_n}2$$
where $H_n=\sum_{k=1}^n\frac1j$ is the $n$-th harmonic number.
The asymptotics of $H_n$ are well-known: $H_n=\ln n+\gamma+O(n^{-1})$.
A: If
$$k=\frac11-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\cdots
$$
then
$$
\frac k2=\frac12-\frac14+\frac16-\frac18+\frac1{10}-\frac1{12}+\frac1{14}-\frac1{16}+\cdots
$$
Determine the sum of these by taking $4$ terms at a time from the first series, and $2$ terms at a time from the second series. For example:
$$
\textstyle\left(\frac11-\frac12+\frac13-\frac14\right)+\left(\frac12-\frac14\right)=\left(\frac11+\frac13\right)+\left(-\frac12+\frac12\right)+\left(-\frac14-\frac14\right)=\frac11-\frac12+\frac13
$$
(The lesson here is that the resulting series is a rearrangement of the first series, but the result sums to a different value.)
