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

$\begin{array}{l}{\mathrm{t}}_{\mathrm{r}}=\frac{\mathrm{r}}{{\mathrm{r}}^4+{\mathrm{r}}^2+1}=\frac{\mathrm{r}}{\left({\mathrm{r}}^2+1\right)-{\mathrm{r}}^2}\\ =\frac{1}{2}\left(\frac{1}{{\mathrm{r}}^2-\mathrm{r}+1}-\frac{1}{{\mathrm{r}}^2+\mathrm{r}+1}\right)\\ =\frac{1}{2}\left(\frac{1}{\mathrm{r}(\mathrm{r}-1)+1}-\frac{1}{(\mathrm{r}+1)\mathrm{r}+1}\right)\\ \therefore \sum _{\mathrm{r}=1}^{\mathrm{n}} {\mathrm{t}}_{\mathrm{r}}=\sum _{\mathrm{r}=1}^{\mathrm{n}} \frac{1}{2}\left[\mathrm{f}\right(\mathrm{r})-\mathrm{f}(\mathrm{r}+1\left)\right]\end{array}$
When 
$\begin{array}{l}\mathrm{f}\left(\mathrm{r}\right)=\frac{1}{\mathrm{r}(\mathrm{r}-1)+1}=\frac{1}{2}\left[\mathrm{f}\right(1)-\mathrm{f}(\mathrm{n}+1\left)\right]\\ =\frac{1}{2}\left(1-\frac{1}{(\mathrm{n}+1)\mathrm{n}+1}\right)\to \frac{1}{2}\text{ as }\mathrm{n}\to \mathrm{\infty }\end{array}$