Question

Find the sum of series 4/(4n-3) (4n+1)

Answer 1

$\dfrac4{(4n-3)(4n+1)}=\dfrac1{4n-3}-\dfrac1{4n-1}$

Assuming the sum starts at $n=1$, the $N$th partial sum is

$\displaystyle\sum_{n=1}^N\left(\frac1{4n-3}-\frac1{4n-1}\right)=\left(1-\frac15\right)+\left(\frac15-\frac19\right)+\cdots+\left(\frac1{4N-7}-\frac1{4N-3}\right)+\left(\frac1{4N-3}-\frac1{4N+1}\right)$
$\displaystyle\sum_{n=1}^N\left(\frac1{4n-3}-\frac1{4n-1}\right)=1-\frac1{4N-1}$

As $N\to\infty$, you're left with simply 1.