Prove that: $1-\frac 12+\frac 13-\frac 14+...+\frac {1}{199}- \frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}$ Prove that:

$$1-\frac 12+\frac 13-\frac 14+...+\frac {1}{199}- \frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}$$

I know only this method: 
$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+....=1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-...$
But, unfortunately, I could not a hint.

$$\sum_{k= 1}^{200}(-1)^{k+1}k^{-1}=1-\frac 12+\frac 13-\frac 14+\cdots+\frac {1}{199}- \frac{1}{200}=\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{200}=$$
$$\color{blue}{\left(1+\frac 13+\frac 15 +\dots+ \frac 1{199}\right)-\left(\frac 12+ \frac 14+\frac 16\cdots+\frac {1}{200}\right)}=\color{red}{\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{200}}$$

\begin{align}
\tag1
\color{blue}{\sum_{k=1}^{100}\frac{1}{2k-1} - \sum_{k=1}^{100}\frac{1}{2k}} &= \color{red}{\sum_{k=101}^{200}\frac{1}{k}}\\
\tag2
&= \sum_{k=1}^{200}\frac{1}{k} - \sum_{k=1}^{100}\frac{1}{k}\\
\tag3
&= \left(\sum_{k=1}^{100}\frac{1}{2k} + \sum_{k=1}^{100}\frac{1}{2k-1}\right) - \sum_{k=1}^{100}\frac{1}{k}\\
\tag4
&= \frac{1}{2}\sum_{k=1}^{100}\frac{1}{k} + \sum_{k=1}^{100}\frac{1}{2k-1}- \sum_{k=1}^{100}\frac{1}{k}\\
\tag5
&= \sum_{k=1}^{100}\frac{1}{2k-1}- \frac{1}{2}\sum_{k=1}^{100}\frac{1}{k}\\
\tag6
&= \underbrace{\color{blue}{\sum_{k=1}^{100}\frac{1}{2k-1}- \sum_{k=1}^{100}\frac{1}{2k}}}\\
&\quad\quad\quad\sum_{k= 1}^{200}(-1)^{k+1}k^{-1}\\
&&\Box
\end{align}

