Proving that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$ I have written the left side of the equation as $$\left(1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}\right).$$ I don't know how to find the sums of these sequences. I know the sums for odd and even integers, but I can't figure this out.

Denote $H_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n}$. What you want to prove is exactly
$$
1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1}-\frac{1}{2n} = H_{2n} - H_n
$$
i.e.,
$$
1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1}-\frac{1}{2n} - H_{2n} = - H_n
$$
Note that the left side is equal to
$$
2(-\frac{1}{2} - \cdots - \frac{1}{2n}) = -H_n
$$
since those terms in odd positions have been removed and those in even positions have been doubled.
