summation - Showing that $1-1/2+ cdots +1/(2n-1)-1/2n$ is equal to $1/(n+1)+1/(n+2)+ cdots +1/(2n)$

$1-1/2+1/3-1/4+ \cdots +1/(2n-1)-1/2n=1/(n+1)+1/(n+2)+ \cdots +1/2n$

I was asked to prove by mathematical induction the validity of the above equation. It isn't hard to prove that it holds for any arbitrary natural number. But how mathematicians (or anyone) discovered that the left side of that equation equals to the right side, it doesn't seem obvious. I've tried to manipulate them with various means such as multiplying the denominator but I can't observe any pattern. Is it by chance that this equation was discovered? Thanks in advance.

Answer

I don't know how someone first discovered this identity, but here's a clearer way of seeing it:

$$\begin{align*} 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots - \frac{1}{2n} &= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2n}- 2\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2n} \right)\\ &= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2n}-\left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \right) \end{align*}$$

This cancels out the first $n$ terms of the sequence, leaving the $(n+1)^\text{st}$ to $2n^\text{th}$ terms, which is the righthand side.