Question

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

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

Answer 1

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

${\sum }_{r=1}^{n}co{t}^{-1}(r^2+\frac{3}{4})={\sum }_{r=1}^{n}ta{n}^{-1}\left(\frac{1}{r^2+\left(\frac{3}{4}\right)}\right)$
$={\sum }_{r=1}^n\frac{(r+\frac{1}{2})-(r-\frac{1}{2})}{1+(r-\frac{1}{2})(r+\frac{1}{2})}$
$={\sum }_{r=1}^n[ta{n}^{-1}(r+\frac{1}{2})-ta{n}^{-1}(r-\frac{1}{2})]$
$=(ta{n}^{-1}\frac{3}{2}-ta{n}^{-1}\frac{1}{2})+(ta{n}^{-1}\frac{5}{2}-ta{n}^{-1}\frac{1}{2})+\cdots +ta{n}^{-1}(n+\frac{1}{2})-ta{n}^{-1}(n-\frac{1}{2})$
$=ta{n}^{-1}(n+\frac{1}{2})-ta{n}^{-1}\frac{1}{2}$
$\therefore \underset{n\to \infty }{lim}{\sum }_{r=1}^{n}co{t}^{-1}(r^2+\frac{3}{4})=ta{n}^{-1}(n+\frac{1}{2})-ta{n}^{-1}\frac{1}{2}$
$=\frac{\pi }{2}-ta{n}^{-1}\frac{1}{2}=co{t}^{-1}\frac{1}{2}=ta{n}^{-1}2$