Prove that $\frac{1}{1*3}+\frac{1}{3*5}+\frac{1}{5*7}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$ Trying to prove that above stated question for $n \geq 1$. A hint given is that you should use $\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1})$. Using this, I think I reduced it to $\frac{1}{2}(\frac{1}{n^2}-(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}))$. Just not sure if it's correct, and what to do with the second half. 

For what it's worth, good ol' induction works, too -- $ \ n = 1 \ $ gives $ \ \frac{1}{3} \ $ and the sum for the $ \ (n+1) \ $ case gives
$$ \frac{n}{2n+1} \ + \ \frac{1}{(2n+1) \ (2n+3)}  \ = \ \frac{n \ (2n+3) \ + \ 1}{(2n+1) \ (2n+3)} \ = \ \frac{  2n^2 + 3n +  1}{(2n+1) \ (2n+3)}  $$
$$= \ \frac{  (2n + 1) \ (n + 1)}{(2n+1) \ (2n+3)} \ = \ \frac{ n + 1}{ 2n+3} \ = \ \frac{(n + 1)}{2(n+1) + 1} \ \ .$$
