Question

$\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{18}+.........+\tan^{-1}\left(\frac{1}{n^2+n+1}\right)+....to\infty$ is equal to

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{4}$
• C. $\frac{2\pi}{3}$
• D. 0

Answer 1

Answer: (B)

$\frac{\pi}{4}$

Answer 2

$T_{n} = tan^{-1} \frac{1}{n^{2}+n+1}$
$= tan^{-1} \frac{1}{1+n\left(n+1\right)}$
$= tan^{-1}\left[\frac{\left(n+1\right)-n}{1+\left(n+1\right)n}\right]$
$= tan^{-1} \left(n+1\right)-tan^{-1}\left(n\right)$
Now $T_{1} = tan^{-1} 2 - tan^{-1}1$
$T_{2} = tan^{-1}3 -tan^{-1}2$
$T_{3} = tan^{-1}4-tan^{-1}3$
$\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots$
$\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots$
$T_{n} = tan^{-1}\left(n+1\right)-tan^{-1} n$
$\therefore T_{1}+T_{2}+..............+T_{n}$
$=tan^{-1}\left(n+1\right)-tan^{-1}1$
$= tan^{-1} \frac{n+1-1}{1+\left(n+1\right)1} = tan^{-1} \frac{n}{n+2}$
Let $n \rightarrow \infty$
$\therefore T_{1}+T_{2}+..........\infty$
$= \displaystyle \lim_{n\rightarrow\infty} \left[tan^{-1} \frac{1}{1+\frac{2}{n}}\right]$
$= tan^{-1}1 = \frac{\pi}{4}$
Hence reqd. sum $= \frac{\pi}{4}$