Question

Prove that $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\dots \dots +\frac{1}{\left(2n-1\right)\left(2n+1\right)}=\frac{n}{2n+1}$

Answer 1

By partial fraction, we know that

$\frac{1}{(2K-1)(2K+1)}=\frac{1}{2}(\frac{1}{2K-1}-\frac{1}{2K+1})$

Using Telescoping sum,

$\sum _{k=1}^n\frac{1}{(2K-1)(2K+1)}=\sum _{k=1}^n\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1})$

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

$=\frac{n}{2n+1}$

Hence, proved.