Is this equality true? Any sources?

Question

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?

Answer 1

Hint: Add $$2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{200}\right)$$to both sides.

Answer 2

$$\sum_{r=n+1}^{2n}\frac1r=\sum_{r=1}^{2n}\frac1r-\sum_{r=1}^n\frac1r$$

$$=\sum_{r=1}^n\frac1{2r-1}+\sum_{r=1}^n\frac1{2r}-\sum_{r=1}^n\frac1r$$

Now $$\sum_{r=1}^n\frac1{2r}-\sum_{r=1}^n\frac1r=\sum_{r=1}^n\left(\frac1{2r}-\frac1r\right)=-\sum_{r=1}^n\frac1{2r}$$

Answer 3

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 $$ \begin{align*} &1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n+1} - \frac{1}{2n+2} \\ = &1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1} - \frac{1}{2n} + \frac{1}{2n+1} - \frac{1}{2n+2} \\ = &\frac{1}{n+1} + \cdots + \frac{1}{2n} + \frac{1}{2n+1} - \frac{1}{2n+2} \\ = &\frac{1}{n+2} + \cdots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{n+1} - \frac{1}{2n+2} \\ = &\frac{1}{n+2} + \cdots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}. \end{align*} $$

Answer 4

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)