Figure out the Master term.!!!(150 followers problem)

This approach might more suitable if asked in some entrance exam!! $\tan ^{ -1 }{ x } +\tan ^{ -1 }{ y } =\tan ^{ -1 }{ \left\{ \frac { x+y }{ 1-xy } \right\} } \\ \therefore \quad \tan ^{ -1 }{ \left( \frac { 2 }{ 4 } \right) } +\tan ^{ -1 }{ \left( \frac { 4 }{ 22 } \right) } =\tan ^{ -1 }{ \left( \frac { 3 }{ 4 } \right) } \quad ...(1)\\ \tan ^{ -1 }{ \left( \frac { 3 }{ 4 } \right) } +\tan ^{ -1 }{ \left( \frac { 6 }{ 92 } \right) } =\tan ^{ -1 }{ \left( \frac { 6 }{ 7 } \right) } \quad ...\quad (2)\\ \tan ^{ -1 }{ \left( \frac { 6 }{ 7 } \right) } +\tan ^{ -1 }{ \left( \frac { 8 }{ 274 } \right) } =\tan ^{ -1 }{ \left( \frac { 10 }{ 11 } \right) } ...(3)\\ From\quad (1),(2)\quad and\quad (3)\quad it\quad can\quad be\quad conclude\quad that\\ If\quad add\quad n\quad terms\quad then\quad final\quad term\quad will\quad be\\ \tan ^{ -1 }{ \left\{ \frac { f(n)\quad -1 }{ f(n) } \right\} } ,\quad Where\quad f(n)\quad is\quad polynomial\quad function\quad of\quad 'n'\\ \lim _{ n\quad \rightarrow \infty }{ \tan ^{ -1 }{ \left\{ \frac { f(n)\quad -1 }{ f(n) } \right\} } } will\quad give\quad \tan ^{ -1 }{ 1 } =\quad \frac { \pi }{ 4 } \quad =\quad 0.785\\ \\ \\$

Let the denominator in each term be of the form $f(n)=an^4+bn^3+cn^2+dn+e\space\space\space (Why?)$ $f(1)=a+b+c+d+e=4$

$f(2)=16a+8b+4c+2d+e=22$

$f(3)=81a +27b+9c+3d+e=92$

$f(4)=256a+64b+16c+4d+e=274$

$f(5)=625a+125b+25c+5d+e=652$

Solving this system of equations we get $a=1,b=0,c=1,d=0,e=2$

$f(n)=n^4+n^2+2$

And numerator is of the form $2n$

$t_n=\tan^{-1}\frac{2n}{n^4+n^2+2}$

$t_n=\tan^{-1}\frac{(n^2+n+1)-(n^2-n+1)}{(n^2+n+1)(n^2-n+1)+1}$

$t_n=\tan^{-1}(n^2+n+1)-\tan^{-1}(n^2-n+1)$

Let S be the sum of infinite terms

Starting from n=1 to n $\rightarrow \infty$

$S=\tan^{-1}(\infty)-\tan^{-1}(1)$

We get $S=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$ $\therefore$ $S=0.785$