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.
 A: My method :

$$\left\{ 1-\frac {1}{2}-\frac {1}{4}-...-\frac {1}{128} \right\}+\left\{ \frac {1}{3}-\frac {1}{6}- \frac{1}{12}-...- \frac{1}{192}\right\}+\left\{\frac {1}{5}-\frac{1}{10}-\frac{1}{20}-...- \frac{1}{160}\right\}+...+\left\{ \frac{1}{99}-\frac{1}{198}\right\}+\left\{ \frac{1}{101}+\frac{1}{103}+\frac{1}{105}+...+\frac{1}{199}\right\}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}$$

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

\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}

A: Hint: prove by induction that $\sum_{i=1}^{2n}\frac{(-1)^{i-1}}{i}=\sum_{i=n+1}^{2n}\frac{1}{i}$.
A: We have 
\begin{split} &&1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{199} - \frac{1}{200} \\&=& 
\left(1 + \frac{1}{3} +  \frac{1}{5} +\cdots + \frac{1}{197}+\frac{1}{199}  \right) - \left(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{200}\right)\\
&=&
 \left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{199} + \frac{1}{200}\right) - 2 \cdot \left(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{200}\right)\\
&=&\left(1 + \frac{1}{2} + \cdots  \frac{1}{100}\right)+\left(  \frac{1}{101}+\cdots +\frac{1}{199} + \frac{1}{200}\right) - \left(\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{100}\right)\\[10pt]
&= &\frac{1}{101} + \frac{1}{102} + \cdots + \frac{1}{200}\end{split}
