Question

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

• A. ${\tan }^{-1}\left(2\right)$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. ${\tan }^{-1}\left(3\right)$

Answer 1

The correct option is A ${\tan }^{-1}\left(2\right)$

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

${\tan }^{-1}\left(\frac{1}{r^2+3/4}\right)$

${\tan }^{-1}\left(\frac{4}{4r^2+3}\right)$

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

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

${\tan }^{-1}\left[\frac{\left(r+\frac{1}{2}\right)-\left(r-\frac{1}{2}\right)}{1+\left(r-\frac{1}{2}\right)\left(r+\frac{1}{2}\right)}\right]$

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

$\left[{\tan }^{-1}\left(\frac{3}{2}\right)-{\tan }^{-1}\left(\frac{1}{2}\right)\right]+\left[{\tan }^{-1}\left(n-\frac{1}{2}\right)-{\tan }^{-1}\left(\frac{3}{2}\right)\right]+\left[{\tan }^{-1}\left(n+\frac{1}{2}\right)-\tan \left(n-\frac{1}{2}\right)\right]$

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

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

$=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{1}{2}\right)$

$=\frac{\pi }{2}-{\cos }^{-1}\left(2\right)$

$={\tan }^{-1}\left(2\right)$

Hence$,$ option $\left(A\right)$ is correct$.$