Summation of the Inverted Tangents

$The\quad summation\quad can\quad be\quad written\quad as\\ \\ \sum _{ m=1 }^{ \infty }{ \tan ^{ -1 }{ \left( \frac { (4{ m }^{ 2 }-2+4m)-(4{ m }^{ 2 }-2-4m) }{ 1+(4{ m }^{ 2 }-2+4m).(4{ m }^{ 2 }-2-4m) } \right) } } \\ \\ Now\quad using\quad the\quad identity\\ \tan ^{ -1 }{ (x) } -\tan ^{ -1 }{ (y) } =\tan ^{ -1 }{ \left( \frac { x-y }{ 1+xy } \right) } \\ \sum _{ m=1 }^{ \infty }{ \left[ \tan ^{ -1 }{ (4{ m }^{ 2 }-2+4m) } -\tan ^{ -1 }{ (4{ m }^{ 2 }-2-4m) } \right] } \\ =\quad [\tan ^{ -1 }{ (6) } -\tan ^{ -1 }{ (-2) } ]+[\tan ^{ -1 }{ (32) } -\tan ^{ -1 }{ (6) } ]\\ \quad \quad ................................-\underset { m\rightarrow \infty }{ lim } \tan ^{ -1 }{ (4{ m }^{ 2 }-2-4m) } \\ Now\quad observe\quad that\quad all\quad the\quad terms\quad except\quad the\quad \\ second\quad and\quad the\quad last\quad ,\quad are\quad being\quad cancelled.\\ Therefore\quad Answer\quad =\quad \boxed { \tan ^{ -1 }{ (2)-\frac { \pi }{ 2 } } }$