The value of $\sum\limits^{3}_{n=1} tan^{-1}\frac{1}{n}$, is
Answer & explanation
Correct answer: option 3
$\sum\limits^{3}_{n=1} tan^{-1}\frac{1}{n}= tan^{-1}+tan^{-1}\frac{1}{2}+tan^{-1}\frac{1}{3}$
$\sum\limits^{3}_{n=1}tan^{-1}\frac{1}{n}=\frac{\pi}{4}+tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}×\frac{1}{3}}\right)$
$\sum\limits^{3}_{n=1}tan^{-1}\frac{1}{n}=\frac{\pi}{4}+tan^{-1}1=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}$