The value of cot left( sumlimits_{r = 1}^inft | vedclass

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(B) Let $S_n = \sum_{r=1}^n 	an^{-1} \left( \frac{4}{4r^2 + 3} ight)$.We can rewrite the argument of the $	an^{-1}$ function as:$\frac{4}{4r^2 + 3

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$\frac{1}{2}$

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(B) Let $S_n = \sum_{r=1}^n 	an^{-1} \left( \frac{4}{4r^2 + 3} ight)$.We can rewrite the argument of the $	an^{-1}$ function as:$\frac{4}{4r^2 + 3} = \frac{(r + 1/2) - (r - 1/2)}{1 + (r + 1/2)(r - 1/2)} = \frac{4}{4r^2 + 1 - 1 + 3} = \frac{4}{4r^2 + 3}$.Thus,$	an^{-1} \left( \frac{4}{4r^2 + 3} ight) = 	an^{-1} \left( r + \frac{1}{2} ight) - 	an^{-1} \left( r - \frac{1}{2} ight)$.This is a telescoping sum:$S_n = \sum_{r=1}^n \left[ 	an^{-1} \left( r + \frac{1}{2} ight) - 	an^{-1} \left( r - \frac{1}{2} ight) ight] = 	an^{-1} \left( n + \frac{1}{2} ight) - 	an^{-1} \left( \frac{1}{2} ight)$.As $n 	o \infty$,$	an^{-1} \left( n + \frac{1}{2} ight) 	o \frac{\pi}{2}$.So,the sum is $\frac{\pi}{2} - 	an^{-1} \left( \frac{1}{2} ight) = \cot^{-1} \left( \frac{1}{2} ight) = 	an^{-1} (2)$.Finally,$\cot \left( 	an^{-1} (2) ight) = \frac{1}{2}$.

The value of $\cot \left( \sum\limits_{r = 1}^\infty \tan^{-1} \left( \frac{4}{4r^2 + 3} \right) \right)$ is equal to

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