Question

Sum to infinite terms of the series $co{t}^{-1}(1^2+\frac{3}{4})+co{t}^{-1}(2^2+\frac{3}{4})+co{t}^{-1}(3^2+\frac{3}{4})+.......$ is

• A. $\frac{\pi }{4}$
• B. $ta{n}^{-1}2$
• C. $ta{n}^{-1}3$
• D. $co{t}^{-1}3$

Answer 1

The correct option is B $ta{n}^{-1}2$

$T_n=co{t}^{-1}(n^2+\frac{3}{4})=ta{n}^{-1}\left(\frac{1}{n^2+\frac{3}{4}}\right)=ta{n}^{-1}(n+\frac{1}{2})-ta{n}^{-1}(n-\frac{1}{2})$
$S_n={\sum }_{n=1}^n{t}_n=ta{n}^{-1}(n+\frac{1}{2})-ta{n}^{-1}\frac{1}{2}\Rightarrow S_{\infty }=\frac{\pi }{2}-ta{n}^{-1}\frac{1}{2}=co{t}^{-1}\frac{1}{2}=Ta{n}^{-1}2$