Question

# The value of $$\displaystyle \sum_{r=1}^{\infty }\tan ^{-1}\frac{2r}{2+r^{2}+r^{4}}$$ is equal to

A
π4
B
π3
C
π2
D
π

Solution

## The correct option is A $$\displaystyle \frac{\pi }{4}$$$$\displaystyle \tan ^{ -1 }{ \frac { 2r }{ { r }^{ 4 }+{ r }^{ 2 }+2 } } =\tan ^{ -1 }{ \frac { 2r }{ 1+\left( { r }^{ 2 }+r+1 \right) \left( { r }^{ 2 }-r+1 \right) } } \\ =\tan ^{ -1 }{ \left( { r }^{ 2 }+r+1 \right) } -\tan ^{ -1 }{ \left( { r }^{ 2 }-r+1 \right) }$$$$\displaystyle \lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \tan ^{ -1 }{ \frac { 2r }{ { r }^{ 4 }+{ r }^{ 2 }+2 } } } } \\ =\lim _{ n\rightarrow \infty }{ \left( \tan ^{ -1 }{ 3 } -\tan ^{ -1 }{ + } \tan ^{ -1 }{ 7 } -\tan ^{ -1 }{ 3 } +...\tan ^{ -1 }{ x } \right) }$$$$\displaystyle =\lim _{ n\rightarrow \infty }{ \left( \tan ^{ -1 }{ \left( \dfrac { { n }^{ 2 }+n }{ { n }^{ 2 }+n+2 } \right) } \right) } =\lim _{ n\rightarrow \infty }{ \left( \tan ^{ -1 }{ \left( \dfrac { 1+\dfrac { 1 }{ n } }{ 1+\dfrac { 1 }{ n } +\dfrac { 2 }{ { n }^{ 2 } } } \right) } \right) }$$$$\displaystyle =\tan ^{ -1 }{ \left( 1 \right) } =\dfrac { \pi }{ 4 }$$Mathematics

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