Question

If $P=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{1}{r+3}\right)Q=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{1}{r+1}\right)R=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{1}{r}\right)$
, then $P-2Q+R$ is

• A. $\underset{x\to \pi }{lim}\frac{-\sin x}{(\pi -x{)}^2}$
• B. $\underset{x\to \pi }{lim}\frac{2\pi (x-\pi {)}^2}{1+\cos x}$
• C. $\sum _{k=1}^3{\cot }^{-1}k$
• D. ${\tan }^{-1}\tan \left(9\right)+{\sin }^{-1}\sin \left(9\right)$

Answer 1

The correct option is D ${\tan }^{-1}\tan \left(9\right)+{\sin }^{-1}\sin \left(9\right)$

$P-2Q+R=\sum _{r=1}^{\infty }{\tan }^{-1}\frac{1}{r+3}-2{\tan }^{-1}\left(\frac{1}{r+1}\right)+{\tan }^{-1}\frac{1}{r}={\tan }^{-1}\frac{1}{4}-2{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{1}+{\tan }^{-1}\frac{1}{5}-2{\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{6}-2{\tan }^{-1}\frac{1}{4}+{\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{7}-2{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{4}⋮⋮⋮⋮⋮⋮$
$={\tan }^{-1}1-{\tan }^{-1}\frac{1}{2}-{\tan }^{-1}\frac{1}{3}=\frac{\pi }{4}-\frac{\pi }{4}=0$
Now
$\text{a.)}L=\underset{x\to \pi }{lim}\frac{-\sin x}{(\pi -x{)}^2}=\underset{x\to \pi }{lim}\frac{-\sin (\pi -x)}{(\pi -x{)}^2}=\underset{h\to 0}{lim}\frac{-\sin h}{h^2}$
As $h\to 0^{-},L\to \infty$
and as $h\to 0^{+},L\to -\infty$

$\text{b.)}\underset{x\to \pi }{lim}\frac{2\pi (x-\pi {)}^2}{1+\cos x}=\underset{x\to \pi }{lim}\frac{4\pi (x-\pi )}{-\sin x}$
$=\underset{x\to \pi }{lim}\frac{4\pi }{-\cos x}=4\pi$

$\text{c.)}\sum _{k=1}^3{\cot }^{-1}k=\frac{\pi }{4}+{\cot }^{-1}2+{\cot }^{-1}3=\frac{\pi }{4}+{\cot }^{-1}\left(\frac{6-1}{3+2}\right)$
$=\frac{\pi }{4}+\frac{\pi }{4}=\frac{\pi }{2}\ne 0$

$\text{d.)}{\tan }^{-1}\tan \left(9\right)+{\sin }^{-1}\sin \left(9\right)=9-3\pi +3\pi -9=0$