Verified Solution • 2025–26 Session moderate QJEEM21Mad4a7: JEE Main 2021 Mathematics - Class 12 JEEM21Mad4a7 Solution limn→∞tan{∑r=1ntan−1(11+r+r2)}\mathop {\lim }\limits_{n \to \infty } \tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {1 + r + {r^2}}}} \right)} } \right\}n→∞limtan{r=1∑ntan−1(1+r+r21)} is equal to ______. Step-by-Step Verified Solution tan−1(11+r+r2){\tan ^{ - 1}}\left( {{1 \over {1 + r + {r^2}}}} \right)tan−1(1+r+r21) =tan−1(r+1−r1+r(r+1))= {\tan ^{ - 1}}\left( {{{r + 1 - r} \over {1 + r(r + 1)}}} \right)=tan−1(1+r(r+1)r+1−r) =tan−1(r+1)−tan−1r= {\tan ^{ - 1}}(r + 1) - {\tan ^{ - 1}}r=tan−1(r+1)−tan−1r ∴\therefore∴ ∑r=1n(tan−1(r+1)−tan−1(r))\sum\limits_{r = 1}^n {\left( {{{\tan }^{ - 1}}(r + 1) - {{\tan }^{ - 1}}(r)} \right)}r=1∑n(tan−1(r+1)−tan−1(r)) =tan−1(2)−tan−1(1)+tan−1(3)−tan1(2)+tan−1(n+1)−tan−1(n)= {\tan ^{ - 1}}(2) - {\tan ^{ - 1}}(1) + ta{n^{ - 1}}(3) - {\tan ^1}(2) + ta{n^{ - 1}}(n + 1) - {\tan ^{ - 1}}(n)=tan−1(2)−tan−1(1)+tan−1(3)−tan1(2)+tan−1(n+1)−tan−1(n) =tan−1(n+1)−tan−1(1)= {\tan ^{ - 1}}(n + 1) - {\tan ^{ - 1}}(1)=tan−1(n+1)−tan−1(1) =tan−1(n+1−11+(n+1)1)= {\tan ^{ - 1}}\left( {{{n + 1 - 1} \over {1 + (n + 1)1}}} \right)=tan−1(1+(n+1)1n+1−1) =tan−1(nn+2)= {\tan ^{ - 1}}\left( {{n \over {n + 2}}} \right)=tan−1(n+2n) limn→∞tan(∑r=1ntan−1(11+r+r2))\mathop {\lim }\limits_{n \to \infty } \tan \left( {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {1 + r + {r^2}}}} \right)} } \right)n→∞limtan(r=1∑ntan−1(1+r+r21)) =limx→∞tan(tan−1(nn+2))= \mathop {\lim }\limits_{x \to \infty } \tan \left( {{{\tan }^{ - 1}}\left( {{n \over {n + 2}}} \right)} \right)=x→∞limtan(tan−1(n+2n)) =limx→∞nn+2= \mathop {\lim }\limits_{x \to \infty } {n \over {n + 2}}=x→∞limn+2n =1= 1=1 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring