Proof by induction: $\sum\limits_{i=1}^{n} \frac{1}{n+i} = \sum\limits_{i=1}^{n} \left(\frac{1}{2i-1} - \frac{1}{2i}\right)$ How can the following be proved by induction? 
$$\sum\limits_{i=1}^{n} \frac{1}{n+i} = \sum\limits_{i=1}^{n} \left(\frac{1}{2i-1} - \frac{1}{2i}\right)$$
I am out of ideas after practicing for a while: 
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots+\left(\frac{1}{2n-1}-\frac{1}{2n}\right)
$$ 
Does this involve telescoping series?

$n=1$, check.  note that
$$
\frac{1}{2(n+1)-1}-\frac{1}{2(n+1)}=\frac{(2n+2)-(2n+1)}{(2n+1)(2n+2)}=\frac{1}{(2n+1)(2n+2)}
$$
(this is what changes on the rhs for $n+1$).
on the left hand side the change is
$$
\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}=\frac{(2n+2)+(2n+1)-2(2n+1)}{(2n+1)(2n+2)}=\frac{1}{(2n+1)(2n+2)}
$$
as desired.
