If the $r^{th}$ term $t_r$ of a series is given by $t_r=\frac{r}{r^4+r^2+1}$ then  $\underset{n\to \infty }{Lt}\sum _{r=1}^n{t}_r$ is

$t_r=\frac{r}{r^4+r^2+1}=\frac{r}{\left(r^2+1\right)-r^2}$

$=\frac{1}{2}\left(\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}\right)$

$=\frac{1}{2}\left(\frac{1}{r\left(r-1\right)+1}-\frac{1}{\left(r+1\right)r+1}\right)$

$\therefore \sum _{r=1}^n{t}_r=\sum _{r=1}^n\frac{1}{2}\left[f\left(r\right)-f\left(r+1\right)\right]$

When  

$f\left(r\right)=\frac{1}{r\left(r-1\right)+1}=\frac{1}{2}\left[f\left(1\right)-f\left(n+1\right)\right]$

$=\frac{1}{2}\left(1-\frac{1}{\left(n+1\right)n+1}\right)\to \frac{1}{2}asn\to \infty$