[tex]\dfrac4{(4n-3)(4n+1)}=\dfrac1{4n-3}-\dfrac1{4n-1}[/tex]
Assuming the sum starts at [tex]n=1[/tex], the [tex]N[/tex]th partial sum is
[tex]\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)[/tex]
[tex]\displaystyle\sum_{n=1}^N\left(\frac1{4n-3}-\frac1{4n-1}\right)=1-\frac1{4N-1}[/tex]
As [tex]N\to\infty[/tex], you're left with simply 1.
