Question

$\underset{n\to \infty }{lim}\sum _{r=1}^{n}co{t}^{-1}(r^2+\frac{3}{4})$ is

• A. 0
• B. tan^-11
• C. tan^-12
• D.

Answer 1

The correct option is C

tan^-12

$\underset{n\to \infty }{lim}\sum _{r=1}^{n}ta{n}^{-1}\left(\frac{4}{4r^2+3}\right)$

= $\underset{n\to \infty }{lim}\sum _{r=1}^{n}ta{n}^{-1}\left(\frac{1}{r^2-\frac{1}{4}+1}\right)$

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

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

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

= $\frac{\pi }{2}-ta{n}^{-1}\frac{1}{2}$

= $ta{n}^{-1}2$