Proof rearranged alternating harmonic series tends to $\frac{3s}2$ The harmonic series $$s:=\sum_{k=1}^\infty (-1)^{k+1}\frac1k$$  tends to $\lim_{n\to \infty}s_n = \ln2$.
This rearrangement of the series is given: $$1+\frac13-\frac12+\frac15+\frac17-\frac14\pm\ldots$$ Show that this series tends to $\frac{3s}2.$
 A: $$
\begin{align}
&\left(1+\frac13-\frac24\right)+\left(\frac15+\frac17-\frac28\right)+\left(\frac19+\frac1{11}-\frac2{12}\right)+\dots\\
&=\left(\color{#C00000}{1-\frac12+\frac13-\frac14}+\color{#00A000}{\frac12-\frac14}\right)+\left(\color{#C00000}{\frac15-\frac16+\frac17-\frac18}+\color{#00A000}{\frac16-\frac18}\right)+\dots\\
&=\left(\color{#C00000}{1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18}+\dots\right)+\left(\color{#00A000}{\frac12-\frac14+\frac16-\frac18}+\dots\right)
\end{align}
$$
The sum of the red part of each group is positive, as is the sum of the green part, thus we can rearrange the red pieces and the green pieces as long as we keep them together. The red part is the normal alternating harmonic series, and the green part is one-half the normal alternating harmonic series.
A: $$ A_n = 1+\frac13-\frac12+\frac15+\frac17-\frac14\pm\ldots$$
thus if you note :
$$ B_n = \sum_{k=0}^n \frac{1}{2k+1} $$
and 
$$ C_n = \sum_{k pair}^n \frac{1}{k} $$
you have $$A_n = B_n - C_n$$
or $S_n = B_n - D_n$ where $D_n = \sum_{k=1}^n \frac{1}{2k}$ 
So just look for a relationship between $D_n$ and $C_n$ and there you go.
A: Let $H_k = \sum_{n=1}^k \frac{1}{n}$
It's well known that $H_k - \ln(k) \rightarrow \gamma$ as $k$ goes to infinity.
Then a partial sum of this is: $H_{4k} - \frac{H_{2k}}{2} - \frac{H_{k}}{2}$
Which, using the approximation for $H_k$ is about $\ln(4k) - \frac{\ln(2k)}{2} - \frac{\ln(k)}{2} = \frac{\ln(2) + \ln(4)}{2} = \frac{3}{2}\ln(2) $ 
