Is this equality true? Any sources? Is it true that $1 - \frac 12+\frac13-\frac14+\cdots-\frac 1{200}=\frac 1{101}+\frac 1{102}+\cdots+\frac 1{200}$? Where can I find sources for this proof?
 A: Hint: Add 
$$2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{200}\right)$$to both sides. 
A: $$\sum_{r=n+1}^{2n}\frac1r=\sum_{r=1}^{2n}\frac1r-\sum_{r=1}^n\frac1r$$
$$=\sum_{r=1}^n\frac1{2r-1}+\sum_{r=1}^n\frac1{2r}-\sum_{r=1}^n\frac1r$$
Now
$$\sum_{r=1}^n\frac1{2r}-\sum_{r=1}^n\frac1r=\sum_{r=1}^n\left(\frac1{2r}-\frac1r\right)=-\sum_{r=1}^n\frac1{2r}$$
A: Here is the boring proof by induction. We want to prove that
$$ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1} - \frac{1}{2n} = \frac{1}{n+1} + \cdots + \frac{1}{2n}. $$
Base case: $n = 1$. The left-hand side reads $1-1/2 = 1/2$, and the right-hand side is $1/2$.
Induction step: Suppose
$$ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1} - \frac{1}{2n} = \frac{1}{n+1} + \cdots + \frac{1}{2n}. $$
Then
$$
\begin{align*}
&1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n+1} - \frac{1}{2n+2} \\ =
&1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1} - \frac{1}{2n} + \frac{1}{2n+1} - \frac{1}{2n+2} \\ =
&\frac{1}{n+1} + \cdots + \frac{1}{2n} + \frac{1}{2n+1} - \frac{1}{2n+2} \\ =
&\frac{1}{n+2} + \cdots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{n+1} - \frac{1}{2n+2} \\ =
&\frac{1}{n+2} + \cdots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}. \end{align*}
$$
A: To make this a bit easier to think about...
$$\sum\limits_{r=1}^{200} \frac{1}{r} = 1 + \frac{1}{2} + \frac{1}{3}+ \cdots + \frac{1}{199} + \frac{1}{200}$$ 
Subtract the even fractions between 1/2 and 1/200:
$$\sum\limits_{r=1}^{200} \frac{1}{r} - \sum\limits_{r=1}^{100} \frac{1}{2r}= 1 + \frac{1}{3} + \cdots + \frac{1}{199}$$ 
Take the even fractions off again
$$\sum\limits_{r=1}^{200} \frac{1}{r} - 2\sum\limits_{r=1}^{100} \frac{1}{2r}= 1 -\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{199} - \frac{1}{200}$$ 
Now,
$$2\sum\limits_{r=1}^{100} \frac{1}{2r} = \sum\limits_{r=1}^{100} \frac{1}{r}$$
So the left-hand side becomes:
$$\sum\limits_{r=1}^{200} \frac{1}{r} - \sum\limits_{r=1}^{100} \frac{1}{r}$$
$$ = \sum\limits_{r=101}^{200} \frac{1}{r}
 = \frac{1}{101} + \frac{1}{102} + \cdots + \frac{1}{200} $$ 
(quod erat demonstrandum)
