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Q.
The value of $\displaystyle\sum_{n=1}^{\infty}\left(\tan ^{-1}\left(\frac{n}{n+2}\right)-\tan ^{-1}\left(\frac{n-1}{n+1}\right)\right)$ is equal to
Inverse Trigonometric Functions
Solution:
We have $T _1=\tan ^{-1} \frac{1}{3}-\tan ^{-1} 0 $
$T _2=\tan ^{-1} \frac{1}{2}-\tan ^{-1} \frac{1}{3} $
$T _3=\tan ^{-1} \frac{3}{5}-\tan ^{-1} \frac{1}{2} $
$\vdots \quad \vdots \quad \vdots$
$T _{ n }=\tan ^{-1}\left(\frac{ n }{ n +2}\right)-\tan ^{-1}\left(\frac{ n -1}{ n +1}\right)$
$\therefore$ On adding all above equation, we get $S _{ n }=\displaystyle\sum_{ r =1}^{ n } T _{ r }=\tan ^{-1}\left(\frac{ n }{ n +2}\right)$
Hence $S_{\infty}=\underset{n \rightarrow \infty}{\text { Limit }} \tan ^{-1}\left(\frac{n}{n+2}\right)=\tan ^{-1} 1=\frac{\pi}{4}$