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Question

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


A
π4
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B
π3
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C
π2
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D
π
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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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