Question

Evaluate: $\underset{x\to \mathrm{\infty }}{lim}tan\left[\sum _{r=1}^{n}ta{n}^{-1}\left(\frac{1}{1+r+r^2}\right)\right]$

Answer 1

Given: $\underset{x\to \mathrm{\infty }}{lim}tan\left[\sum _{r=1}^{n}ta{n}^{-1}\left(\frac{1}{1+r+r^2}\right)\right]$

Using the identity

${\tan }^{-1}A-{\tan }^{-1}B={\tan }^{-1}\left(\frac{A-B}{1+AB}\right)$

$\Rightarrow \underset{x\to \mathrm{\infty }}{lim}tan\left[\sum _{r=1}^{n}ta{n}^{-1}\left(\frac{\left(r+1\right)-r}{1+r+r^2}\right)\right]$

Let $A=r+1\&b=r$

$$\begin{gathered} \Rightarrow \underset{x\to \mathrm{\infty }}{lim}tan\left[\sum _{r=1}^n\left\{{\tan }^{-1}(r+1)-{\tan }^{-1}\left(r\right)\right\}\right] \\ \Rightarrow \underset{x\to \mathrm{\infty }}{lim}tan\left[{\tan }^{-1}\left(2\right)-{\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(3\right)-{\tan }^{-1}\left(2\right)+{\tan }^{-1}\left(4\right)+........+{\tan }^{-1}\left(n\right)-{\tan }^{-1}\left(n-1\right)+{\tan }^{-1}\left(n+1\right)-{\tan }^{-1}\left(n\right)\right] \\ \Rightarrow \underset{x\to \mathrm{\infty }}{lim}tan\left[{\tan }^{-1}\left(n+1\right)-{\tan }^{-1}\left(1\right)\right] \\ \Rightarrow tan\left[{\tan }^{-1}\left(\infty \right)-{\tan }^{-1}\left(1\right)\right] \\ \Rightarrow tan\left[{\tan }^{-1}\tan \left(\frac{\mathrm{\pi }}{2}\right)-{\tan }^{-1}\tan \left(\frac{\mathrm{\pi }}{4}\right)\right] \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow 1 \end{gathered}$$Therefore,

$\underset{x\to \mathrm{\infty }}{lim}tan\left[\sum _{r=1}^{n}ta{n}^{-1}\left(\frac{1}{1+r+r^2}\right)\right]$$=1$