JEE Main 2025 Apr 4 Shift 2, Mathematics Q3: Summation of Series

The arguments are $\frac{7}{4},\frac{19}{4},\frac{39}{4},\frac{67}{4},\dots$ The numerators $7,19,39,67,\dots$ fit $4n^2+3$ for $n=1,2,3,\dots$. So

$T_n={\cot }^{-1}\left(\frac{4n^2+3}{4}\right)={\tan }^{-1}\left(\frac{4}{4n^2+3}\right)$.

Write $4n^2+3=4\left[\left(n+\frac{1}{2}\right)\left(n-\frac{1}{2}\right)+1\right]$, so

$T_n={\tan }^{-1}\left(\frac{(n+\frac{1}{2})-(n-\frac{1}{2})}{1+(n+\frac{1}{2})(n-\frac{1}{2})}\right)={\tan }^{-1}\left(n+\frac{1}{2}\right)-{\tan }^{-1}\left(n-\frac{1}{2}\right)$.

This telescopes:

$\sum _{n=1}^N{T}_n={\tan }^{-1}\left(N+\frac{1}{2}\right)-{\tan }^{-1}\left(\frac{1}{2}\right)$.

As $N\to \infty$, ${\tan }^{-1}(N+\frac{1}{2})\to \frac{\pi }{2}$. Hence the sum is $\frac{\pi }{2}-{\tan }^{-1}\left(\frac{1}{2}\right)$.