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)

$T_n=ta{n}^{-1}\frac{1}{n^2+n+1}$
$=ta{n}^{-1}\frac{1}{1+n\left(n+1\right)}$
$=ta{n}^{-1}\left[\frac{\left(n+1\right)-n}{1+\left(n+1\right)n}\right]$
$=ta{n}^{-1}\left(n+1\right)-ta{n}^{-1}\left(n\right)$
Now $T_1=ta{n}^{-1}2-ta{n}^{-1}1$
$T_2=ta{n}^{-1}3-ta{n}^{-1}2$
$T_3=ta{n}^{-1}4-ta{n}^{-1}3$
$\dots \dots \dots \dots \dots \dots \dots \dots \dots$
$\dots \dots \dots \dots \dots \dots \dots \dots \dots$
$T_n=ta{n}^{-1}\left(n+1\right)-ta{n}^{-1}n$
$\therefore T_1+T_2+..............+T_n$
$=ta{n}^{-1}\left(n+1\right)-ta{n}^{-1}1$
$=ta{n}^{-1}\frac{n+1-1}{1+\left(n+1\right)1}=ta{n}^{-1}\frac{n}{n+2}$
Let $n\to \infty$
$\therefore T_1+T_2+..........\infty$
$=\underset{n\to \infty }{\lim }\left[ta{n}^{-1}\frac{1}{1+\frac{2}{n}}\right]$
$=ta{n}^{-1}1=\frac{\pi }{4}$
Hence reqd. sum $=\frac{\pi }{4}$