<s>[INST] 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?
 [/INST] 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)
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