Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

$S_n=\sum _{r=1}^n{\tan }^{-1}\left(\frac{2(2r-1)}{4+r^2\left(r^2-2r+1\right)}\right)$

$=\sum _{r=1}^n{\tan }^{-1}\left(\frac{2(2r-1)}{4+r^2(r-1{)}^2}\right)$

$=\sum _{r=1}^n{\tan }^{-1}\left(\frac{\frac{2r-1}{2}}{1+\frac{r^2}{2}\frac{(r-1{)}^2}{2}}\right)$

$=\sum _{r=1}^n{\tan }^{-1}\left(\frac{\frac{{\mathrm{r}}^2}{2}-\frac{(\mathrm{r}-1{)}^2}{2}}{1+\frac{{\mathrm{r}}^2}{2}\cdot \frac{(\mathrm{r}-1{)}^2}{2}}\right)$

$=\sum _{r=1}^n\left({\tan }^{-1}\frac{{\mathrm{r}}^2}{2}-{\tan }^{-1}\frac{(\mathrm{r}-1{)}^2}{2}\right)$

${\mathrm{S}}_{\mathrm{n}}={\tan }^{-1}\frac{{\mathrm{n}}^2}{2}\Rightarrow \cot \left({\mathrm{S}}_{\mathrm{n}}\right)=\frac{2}{{\mathrm{n}}^2}$

${\mathrm{S}}_{\mathrm{n}-1}={\tan }^{-1}\left(\frac{(\mathrm{n}-1{)}^2}{2}\right)\Rightarrow \cot \left({\mathrm{S}}_{\mathrm{n}-1}\right)=\frac{2}{(\mathrm{n}-1{)}^2}$

$\begin{array}{rlrlr}& \cot \left(S_{n-1}\right)-\cot \left(S_n\right)=\frac{2}{(n-1{)}^2}-\frac{2}{n^2}& & & \end{array}$

$\underset{n\to \infty }{\lim }{\sum }_{n=2}^n\left(\cot \left(S_n\right)-\cot S_n\right)=\underset{n\to \infty }{\lim }{\sum }_{n=2}^n\left(\frac{2}{(n-1{)}^2}-\frac{2}{n^2}\right)$

$=\underset{n\to \infty }{\lim }\left[\left(\frac{2}{1^2}-\frac{2}{2^2}\right)+\left(\frac{2}{2^2}-\frac{2}{3^2}\right)+\left(\frac{2}{3^2}-\frac{2}{4^2}\right)+\left(\frac{2}{(n-1{)}^2}-\frac{2}{n^2}\right)\right]$

$=\underset{n\to \infty }{\lim }\left(2-\frac{2}{n^2}\right)=2$