The value of $tan^{-1}\frac{1}{3}+tan^{-1}\frac{1}{7}+tan^{-1}\frac{1}{13}+..........+tan^{-1}\frac{1}{n^2+n+1}+ ....$ to ∞ , is
Answer & explanation
Correct answer: option 2
We have,
$tan^{-1}\frac{1}{3}+tan^{-1}\frac{1}{7}+tan^{-1}\frac{1}{13}+..........+tan^{-1}\frac{1}{n^2+n+1}+ ....$ to ∞
$=\lim\limits_{n→∞} \sum\limits^{n}_{r=1}tan^{-1}\frac{1}{r^2+r+1}$
$=\lim\limits_{n→∞} \sum\limits^{n}_{r=1}tan^{-1} \begin{Bmatrix}\frac{1}{1+r(r+1)}\end{Bmatrix}$
$=\lim\limits_{n→∞} \sum\limits^{n}_{r=1}tan^{-1} \begin{Bmatrix}\frac{(r+1)-r}{1+r(r+1)}\end{Bmatrix}$
$=\lim\limits_{n→∞} \sum\limits^{n}_{r=1} \begin{Bmatrix} tan^{-1}(r+1) -tan^{-1}r \end{Bmatrix}$
$=\lim\limits_{n→∞} \begin{Bmatrix} tan^{-1}(n+1) -tan^{-1} 1\end{Bmatrix}$
$= tan^{-1} ∞ -tan^{-1} 1 = \frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$