Prove that $\frac{1}{1*3}+\frac{1}{3*5}+\frac{1}{5*7}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$

Question

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.

Answer 1

Hint: Multiply and divide by $2$.

What you have done is correct. All you need to observe is terms cancel out in pairs starting from $2nd$ term.

$$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...)$$

So we are left with $1st$ and the last terms. $$\frac{1}{2}(1-\frac{1}{2n+1})$$ $$\frac{n}{2n+1}$$

Answer 2

$$\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1}) $$This implies that the sum is $$\frac{1}{2}\big\{\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{2n-1}+\frac{1}{2n-1}-\frac{1}{2n+1}\big\}$$ cancel terms and complete

Answer 3

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

$$ \frac{n}{2n+1} \ + \ \frac{1}{(2n+1) \ (2n+3)} \ = \ \frac{n \ (2n+3) \ + \ 1}{(2n+1) \ (2n+3)} \ = \ \frac{ 2n^2 + 3n + 1}{(2n+1) \ (2n+3)} $$

$$= \ \frac{ (2n + 1) \ (n + 1)}{(2n+1) \ (2n+3)} \ = \ \frac{ n + 1}{ 2n+3} \ = \ \frac{(n + 1)}{2(n+1) + 1} \ \ .$$