for every integer $n \ge 1$ one has the equality Could any one help me?
For every integer $n \ge 1$ one has the equality:
$$ 1-{1\over 2}+ {1\over 3}-{1\over 4}+\dots+{1\over 2n-1}-{1\over 2n}={1\over n+1}+{1\over n+2}+\dots +{1\over 2n}.$$
Should I proceed by Induction?
 A: Here's another way to do the problem,$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{2n}\\=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2n}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}\right)\\=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2n}-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)\\=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$
If I am not mistaken,this is the Catalan's Identity.
A: Sorry, Prism, I didn't see your answer before I posted this.
Yes, induction works. Check whether it is true for $n=1$. Then, assume that it is true for some $n\in\mathbb{Z}_+$. Now, observe that when you move from $n$ to $n+1$, all you really need to do is add
$\frac{1}{2n+1}-\frac{1}{2n+2}$
to the left-hand side and
$-\frac{1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}$
to the right-hand side. Establish that these two quantities are equal and conclude that the result holds for $n+1$ as well.
A: Yes, one way would be by induction. In the inductive hypothesis, you will need to prove
$$\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{n+1}=\frac{1}{2n+1}+\frac{1}{2n+2}$$
which is clearly true.
