Looks like the series is supposed to be
[tex]\dfrac{13}{2\times3}+\dfrac{13}{3\times4}+\dfrac{13}{4\times5}+\cdots+\dfrac{13}{(n+1)(n+2)}[/tex]
or in condensed form,
[tex]\displaystyle\sum_{k=1}^n\frac{13}{(k+1)(k+2)}[/tex]
Decompose the summand into partial fractions:
[tex]\dfrac{13}{(k+1)(k+2)}=\dfrac a{k+1}+\dfrac b{k+2}[/tex]
[tex]\implies 13=a(k+2)+b(k+1)[/tex]
[tex]k=-1\implies13=a[/tex]
[tex]k=-2\implies13=-b\implies b=-13[/tex]
So the [tex]n[/tex]th partial sum of the series is
[tex]S_n=\displaystyle\sum_{k=1}^n\frac{13}{k+1}-\frac{13}{k+2}[/tex]
[tex]\displaystyle S_k=\left(\frac{13}2-\frac{13}3\right)+\left(\frac{13}3-\frac{13}4\right)+\cdots+\left(\frac{13}n-\frac{13}{n+1}\right)+\left(\frac{13}{n+1}-\frac{13}{n+2}\right)[/tex]
[tex]\implies S_k=\dfrac{13}2-\dfrac{13}{n+2}[/tex]
As [tex]n\to\infty[/tex], the second term converges to 0, leaving the value of the infinite series,
[tex]\displaystyle\sum_{k=1}^\infty\frac{13}{(k+1)(k+2)}=\boxed{\frac{13}2}[/tex]
