The value of sumlimits_{r = 1}^infty {{{tan | vedclass

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(C) Let $T_r = 	an^{-1} \left( \frac{3}{r^2 - r + 9} ight)$.We can rewrite the argument as $\frac{\frac{r}{3} - \frac{r-1}{3}}{1 + \frac{r}{3} \cdo

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$\frac{\pi }{2}$

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(C) Let $T_r = 	an^{-1} \left( \frac{3}{r^2 - r + 9} ight)$.We can rewrite the argument as $\frac{\frac{r}{3} - \frac{r-1}{3}}{1 + \frac{r}{3} \cdot \frac{r-1}{3}} = \frac{3}{r^2 - r + 9}$.Thus,$T_r = 	an^{-1} \left( \frac{r}{3} ight) - 	an^{-1} \left( \frac{r-1}{3} ight)$.This is a telescoping series.The sum $S_n = \sum_{r=1}^n T_r = \left( 	an^{-1} \frac{1}{3} - 	an^{-1} 0 ight) + \left( 	an^{-1} \frac{2}{3} - 	an^{-1} \frac{1}{3} ight) + \dots + \left( 	an^{-1} \frac{n}{3} - 	an^{-1} \frac{n-1}{3} ight)$.$S_n = 	an^{-1} \left( \frac{n}{3} ight) - 	an^{-1} 0 = 	an^{-1} \left( \frac{n}{3} ight)$.Taking the limit as $n 	o \infty$,$S = \lim_{n 	o \infty} 	an^{-1} \left( \frac{n}{3} ight) = \frac{\pi}{2}$.

The value of $\sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\frac{3}{{{r^2} - r + 9}}} \right)} $ is-

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