66Rewrite the r-th term as
t_r=\dfrac{r}{(r^2+1)^2-r^2}=\dfrac{1}{2}\dfrac{(r^2+r+1)-(r^2-r+1)}{(r^2+r+1)(r^2-r+1)}
which gives
t_r=\dfrac{1}{2}\left(\dfrac{1}{r^2-r+1}-\dfrac{1}{r^2+r+1}\right)
Now you can telescope to get the required sum as 1/2
le the rth term tr of a series is given by t_{r}=\frac{r}{1+r^{2}+r^{4}}
then find the value of
\lim_{n\rightarrow \infty }\sum_{r=1}^{n}{t}_{r}
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66