Question

If $f\left(n\right)={\cot }^{-1}(n+3)-2{\cot }^{-1}(n+1)+{\cot }^{-1}(n-1)$ for $n\in N,$ then $\sum _{n=1}^{\infty }f\left(n\right)$ is equal to

• A. $\sum _{k=1}^3{\tan }^{-1}\left(\frac{1}{k}\right)$
• B. $\underset{x\to \pi }{lim}\frac{-\sin \left(x\right)}{\pi -x}$
• C. $\underset{x\to 0^{-}}{lim}{\cot }^{-1}\left(\frac{1}{x}\right)$
• D. $\underset{x\to \pi }{lim}\frac{\pi (x-\pi {)}^2}{4(\cos x+1)}$

Answer 1

Answer: (A), (D)

$\underset{x\to \pi }{lim}\frac{\pi (x-\pi {)}^2}{4(\cos x+1)}$

$f\left(n\right)={\cot }^{-1}(n+3)-2{\cot }^{-1}(n+1)+{\cot }^{-1}(n-1)$
As ${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2},$
$f\left(n\right)={\tan }^{-1}(n+1)-{\tan }^{-1}(n-1)-\left({\tan }^{-1}\right(n+3)-{\tan }^{-1}(n+1\left)\right)$

$f\left(1\right)={\tan }^{-1}2-{\tan }^{-1}0-({\tan }^{-1}4-{\tan }^{-1}2)$
$f\left(2\right)={\tan }^{-1}3-{\tan }^{-1}1-({\tan }^{-1}5-{\tan }^{-1}3)$
$f\left(3\right)={\tan }^{-1}4-{\tan }^{-1}2-({\tan }^{-1}6-{\tan }^{-1}4)$
$f\left(4\right)={\tan }^{-1}5-{\tan }^{-1}3-({\tan }^{-1}7-{\tan }^{-1}5)$
$\cdot \cdot$
$\cdot \cdot$
$\cdot \cdot$
$f(n-1)={\tan }^{-1}\left(n\right)-{\tan }^{-1}(n-2)-\left({\tan }^{-1}\right(n+2)-{\tan }^{-1}(n\left)\right)$
$f\left(n\right)={\tan }^{-1}(n+1)-{\tan }^{-1}(n-1)-\left({\tan }^{-1}\right(n+3)-{\tan }^{-1}(n+1\left)\right)$

$\sum f\left(n\right)=f\left(1\right)+f\left(2\right)+\cdots +f(n-1)+f\left(n\right)={\tan }^{-1}2+{\tan }^{-1}3-{\tan }^{-1}0-{\tan }^{-1}1+{\tan }^{-1}\left(n\right)+{\tan }^{-1}(n+1)-{\tan }^{-1}(n+2)-{\tan }^{-1}(n+3)$
Since $2,3>0,2\times 3>0$
$\therefore \sum _{n=1}^{\infty }f\left(n\right)=\pi +{\tan }^{-1}\left(\frac{2+3}{1-2\times 3}\right)-\frac{\pi }{4}\Rightarrow \sum _{n=1}^{\infty }f\left(n\right)=\frac{\pi }{2}$

$\sum _{k=1}^3{\tan }^{-1}\left(\frac{1}{k}\right)=\frac{\pi }{4}+{\cot }^{-1}2++{\cot }^{-1}3=\frac{\pi }{4}+\frac{\pi }{2}-{\tan }^{-1}2+\frac{\pi }{2}-{\tan }^{-1}3=\pi +\frac{\pi }{4}-(\pi +{\tan }^{-1}\left(\frac{2+3}{1-2\times 3}\right))=\frac{\pi }{2}$

$\underset{x\to \pi }{lim}\frac{-\sin \left(x\right)}{\pi -x}=\underset{x\to \pi }{lim}\frac{-\sin (\pi -x)}{\pi -x}$
Let $x-\pi =t$
As $x\to \pi ,$ $t\to 0$
$\Rightarrow \underset{t\to 0}{lim}\frac{-\sin t}{t}=-1$

$\underset{x\to 0^{-}}{lim}{\cot }^{-1}\left(\frac{1}{x}\right)$
$=\underset{x\to 0^{-}}{lim}(\pi +{\tan }^{-1}\left(x\right))=\pi$ $(\because {\tan }^{-1}\frac{1}{x}=-\pi +{\cot }^{-1}x\text{for}x<0)$

$\underset{x\to \pi }{lim}\frac{\pi (x-\pi {)}^2}{4(\cos x+1)}$
$=\underset{x\to \pi }{lim}\frac{\pi (x-\pi {)}^2}{4(-\cos (\pi -x)+1)}$
Let $x-\pi =y$
Then $\underset{y\to 0}{lim}\frac{\pi (y{)}^2}{4(-\cos y+1)}$
$=\underset{y\to 0}{lim}\frac{\pi (y{)}^2}{8{\sin }^2\frac{y}{2}}=\frac{\pi }{2}$