I'm guessing the sum is supposed to be
[tex]\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}[/tex]
Split the summand into partial fractions:
[tex]\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}[/tex]
[tex]1=a(5k+4)+b(5k-1)[/tex]
If [tex]k=-\frac45[/tex], then
[tex]1=b(-4-1)\implies b=-\frac15[/tex]
If [tex]k=\frac15[/tex], then
[tex]1=a(1+4)\implies a=\frac15[/tex]
This means
[tex]\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}[/tex]
Consider the [tex]n[/tex]th partial sum of the series:
[tex]S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)[/tex]
The sum telescopes so that
[tex]S_n=\dfrac2{14}-\dfrac2{5n+4}[/tex]
and as [tex]n\to\infty[/tex], the second term vanishes and leaves us with
[tex]\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}=\lim_{n\to\infty}S_n=\frac17[/tex]
