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.
My method :
$$\left\{ 1-\frac {1}{2}-\frac {1}{4}-...-\frac {1}{128} \right\}+\left\{ \frac {1}{3}-\frac {1}{6}- \frac{1}{12}-...- \frac{1}{192}\right\}+\left\{\frac {1}{5}-\frac{1}{10}-\frac{1}{20}-...- \frac{1}{160}\right\}+...+\left\{ \frac{1}{99}-\frac{1}{198}\right\}+\left\{ \frac{1}{101}+\frac{1}{103}+\frac{1}{105}+...+\frac{1}{199}\right\}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}$$
Following @Daniel Fischer's comment: Write $-\frac{1}{2k}$ as $+\frac{1}{2k}-\frac{1}{k}$
we obtain \begin{align*} \color{blue}{\sum_{j=1}^{200}(-1)^{j+1}\frac{1}{j}}&=\sum_{j=1}^{100}\frac{1}{2j-1}-\sum_{j=1}^{100}\frac{1}{2j}\\ &=\sum_{j=1}^{100}\frac{1}{2j-1}+\left(\sum_{j=1}^{100}\frac{1}{2j}-\sum_{j=1}^{100}\frac{1}{j}\right)\\ &=\sum_{j=1}^{200}\frac{1}{j}-\sum_{j=1}^{100}\frac{1}{j}\\ &\color{blue}{=\sum_{j=101}^{200}\frac{1}{j}} \end{align*}
Note: With respect to a comment from OP a small supplement to @Zaharyas nice approach.
We consider a smaller sum to better see what's going on. The following is valid \begin{align*} 1-\frac{1}{2}& +\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{13}-\frac{1}{14}+\frac{1}{15}-\frac{1}{16}\\ &=\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}\tag{1} \end{align*}
The main idea is to reorder the left-hand side of (1) and organise it in blocks. Each block starts with a fraction with an odd denominator as leader and we put all numbers within a block where the denominator is a power of two of this odd denominator.
We obtain \begin{align*} 1-\frac{1}{2}& +\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{13}-\frac{1}{14}+\frac{1}{15}-\frac{1}{16}\\ &=\underbrace{\left(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}\right)}_{\color{blue}{=\frac{1}{16}}} +\underbrace{\left(\frac{1}{3}-\frac{1}{6}-\frac{1}{12}\right)}_{\color{blue}{=\frac{1}{12}}}\\ &\qquad+\underbrace{\left(\frac{1}{5}-\frac{1}{10}\right)}_{\color{blue}{=\frac{1}{10}}} +\underbrace{\left(\frac{1}{7}-\frac{1}{14}\right)}_{\color{blue}{=\frac{1}{14}}}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\frac{1}{15}\tag{2}\\ &=\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}\tag{3} \end{align*}
Comment:
In (2) we observe the right-hand side is organized into $8$ blocks. The first four blocks contain more than one number and four blocks which consist of one number only. We note that each natural number $a\in \mathbb{N}=\{1,2,3,\ldots\}$ has a unique representation as power of two times an odd number. \begin{align*} a=b\cdot 2^k\qquad\qquad k\geq 0, b\text{ odd} \end{align*} The second block in the right-hand side of (2) for instance has the representation \begin{align*} \left(\frac{1}{3}-\frac{1}{6}-\color{blue}{\frac{1}{12}}\right)&=\left(\frac{1}{3\cdot 2^0}-\frac{1}{3\cdot2^1}-\frac{1}{3\cdot2^2}\right) =\frac{1}{3}\left(1-\frac{1}{2}-\frac{1}{4}\right)\\ &=\frac{1}{3}\cdot\frac{1}{4} \color{blue}{=\frac{1}{12}} \end{align*} This unique representation asserts that we take all fractions $\frac{1}{k}$ with $1\leq k\leq 16$ in the right-hand side of (2).
Another nice fact is that the value of each block is given by its right-most entry. This is due to the finite geometric series formula. We obtain \begin{align*} 1-\frac{1}{2}-\frac{1}{4}-\cdots\color{blue}{-\frac{1}{2^{k}}} &=1-\frac{1}{2}\left(1+\frac{1}{2}+\cdots+\frac{1}{2^{k-1}}\right)\\ &=1-\frac{1}{2}\sum_{j=0}^{k-1}\frac{1}{2^j}\\ &=1-\frac{1}{2}\cdot\frac{1-\frac{1}{2^k}}{1-\frac{1}{2}}\\ &\color{blue}{=\frac{1}{2^k}} \end{align*}
In (3) we write for each block the corresponding fraction and reorder the numbers.
Keeping all this in mind we can now show the original identity using sigma notation. We obtain \begin{align*} \color{blue}{\sum_{j=1}^{200}(-1)^{j+1}\frac{1}{j}}& =\sum_{j=1}^{50}\left(\frac{1}{2j-1}-\sum_{k=1}^{\left\lfloor\log_{2}\left(\frac{200}{2j-1}\right)\right\rfloor}\frac{1}{(2j-1)\cdot 2^k}\right) +\sum_{j=51}^{100}\frac{1}{2j-1}\tag{4}\\ &=\sum_{j=1}^{50}\frac{1}{(2j-1)\cdot 2^{\left\lfloor\log_{2}\left(\frac{200}{2j-1}\right)\right\rfloor}}+\sum_{j=51}^{100}\frac{1}{2j-1}\tag{5}\\ &=\sum_{j=51}^{100}\frac{1}{2j}+\sum_{j=51}^{100}\frac{1}{2j-1}\tag{6}\\ &\color{blue}{=\sum_{j=101}^{200}\frac{1}{j}} \end{align*}
Comment:
In (4) we use the fact that we have $200$ summands, $100$ with odd denominator and $100$ with even denominator. We take the $100$ odd as leaders of $100$ blocks. Thereby we have $50$ blocks with more than one term and $50$ block with one term only. This corresponds to the representation in (2).
In (5) we use that each block is given by the right-most term as indicated by braces in (2) and also shown in the comment-section above. We can verify the expression $(2j-1)\cdot 2^{\left\lfloor\log_{2}\left(\frac{200}{2j-1}\right)\right\rfloor}$ with $$\text{Sort[Table[(2*j-1)*2^Floor[ld(200/(2*j-1))],{j,1,50}]]}$$ in Wolfram Alpha and get \begin{align*} 102,104,106,\ldots,198,200 \end{align*}
In (6) we use the unique representation of even numbers as odd number times a power of two and collect them in sorted order as we did in (3).
$$\sum_{k= 1}^{200}(-1)^{k+1}k^{-1}=1-\frac 12+\frac 13-\frac 14+\cdots+\frac {1}{199}- \frac{1}{200}=\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{200}=$$ $$\color{blue}{\left(1+\frac 13+\frac 15 +\dots+ \frac 1{199}\right)-\left(\frac 12+ \frac 14+\frac 16\cdots+\frac {1}{200}\right)}=\color{red}{\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{200}}$$
\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}
Hint: prove by induction that $\sum_{i=1}^{2n}\frac{(-1)^{i-1}}{i}=\sum_{i=n+1}^{2n}\frac{1}{i}$.
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}
Just discovered a neat solution:
Add $ 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{100} $ to both sides and it will be easy to see they are indeed equal.
Because on LHS, every negative term can be added with a positive term twice its size. Examples $$-\frac{1}{2} + 1 = \frac{1}{2}$$ $$-\frac{1}{4} + \frac{1}{2}= \frac{1}{4}$$ $$-\frac{1}{6} + \frac{1}{3}= \frac{1}{6}$$
and so on.
