Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

The general term of the series is $T_n={\tan }^{-1}\left(\frac{1}{n^2+3n+3}\right)$. We can rewrite the argument as $\frac{1}{1+(n^2+3n+2)}$. Factoring the quadratic term, $n^2+3n+2=(n+1)(n+2)$. So, $T_n={\tan }^{-1}\left(\frac{1}{1+(n+1)(n+2)}\right)$. We can express the numerator $1$ as $(n+2)-(n+1)$. Using the identity ${\tan }^{-1}A-{\tan }^{-1}B={\tan }^{-1}\left(\frac{A-B}{1+AB}\right)$, we get $T_n={\tan }^{-1}(n+2)-{\tan }^{-1}(n+1)$. The partial sum $S_N={\sum }_{n=1}^N({\tan }^{-1}(n+2)-{\tan }^{-1}(n+1))$ is a telescoping sum: $S_N=({\tan }^{-1}3-{\tan }^{-1}2)+({\tan }^{-1}4-{\tan }^{-1}3)+\cdots +({\tan }^{-1}(N+2)-{\tan }^{-1}(N+1))={\tan }^{-1}(N+2)-{\tan }^{-1}2$. Taking the limit as $N\to \infty$, the sum is ${\lim }_{N\to \infty }({\tan }^{-1}(N+2)-{\tan }^{-1}2)=\frac{\pi }{2}-{\tan }^{-1}2$. Since $\frac{\pi }{2}-{\tan }^{-1}x={\cot }^{-1}x$, the sum is ${\cot }^{-1}2$.