Let the $${r^{th}}$$ term, $${t_r},$$ of a series is given by $${t_r} = \frac{r}{{1 + {r^2} + {r^4}}}.$$     Then $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{t_r}} $$   is :

Solution :
$$\eqalign{ & {t_r} = \frac{1}{2}.\frac{{2r}}{{{{\left( {{r^2} + 1} \right)}^2} - {r^2}}} \cr & = \frac{1}{2}\left\{ {\frac{1}{{{r^2} - r + 1}} - \frac{1}{{{r^2} + r + 1}}} \right\} \cr & = \frac{1}{2}\left\{ {\frac{1}{{r\left( {r - 1} \right) + 1}} - \frac{1}{{\left( {r + 1} \right)r + 1}}} \right\} \cr & \therefore \sum\limits_{r = 1}^n {{t_r}} = \sum\limits_{r = 1}^n {\frac{1}{2}\left\{ {f\left( r \right) - f\left( {r + 1} \right)} \right\}} {\text{, where }}f\left( r \right) = \frac{1}{{r\left( {r - 1} \right) + 1}} \cr & = \frac{1}{2}\left\{ {f\left( 1 \right) - f\left( {n + 1} \right)} \right\} \cr & = \frac{1}{2}\left\{ {1 - \frac{1}{{\left( {n + 1} \right)n + 1}}} \right\} \to \frac{1}{2}{\text{ as }}n \to \infty \cr} $$