Series Exercise Converges to 1

Exercise:

Show that, $\sum\limits_{n=1}^{\infty }{\frac{4}{\left( 4n-3 \right)\left( 4n+1 \right)}=1}$

Solution:  we have $\frac{4}{\left( 4n-3 \right)\left( 4n+1 \right)}=\frac{A}{4n-3}+\frac{B}{4n+1}$

So $4A+4B=0\,\,\,\And \,\,\,A-3B=4\Leftrightarrow A=1\,\,\And \,\,B=-1$

Hence $\sum\limits_{n=1}^{\infty }{\frac{4}{\left( 4n-3 \right)\left( 4n+1 \right)}=\sum\limits_{n=1}^{\infty }{\frac{1}{4n-3}-\sum\limits_{n=1}^{\infty }{\frac{1}{4n+1}}}}=1+\sum\limits_{n=1}^{\infty }{\frac{1}{4\left( n+1 \right)-3}-\sum\limits_{n=1}^{\infty }{\frac{1}{4n+1}}}$

$\sum\limits_{n=1}^{\infty }{\frac{4}{\left( 4n-3 \right)\left( 4n+1 \right)}}=1+\sum\limits_{n=1}^{\infty }{\frac{1}{4n+4-3}-\sum\limits_{n=1}^{\infty }{\frac{1}{4n+1}=1+\sum\limits_{n=1}^{\infty }{\frac{1}{4n+1}-\sum\limits_{n=1}^{\infty }{\frac{1}{4n+1}=1}}}}$