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  4. Find the sum of the series tan -1 (1/2)+ tan -1 (1/8)+ tan -1 (1/18)+ tan -1 (1/32)+ ldots . .+∞ is

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Q. Find the sum of the series $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{8}+\tan ^{-1} \frac{1}{18}+\tan ^{-1} \frac{1}{32}+\ldots . .+\infty$ is

Inverse Trigonometric Functions

Solution:

We have,
$\tan ^{-1}\left(\frac{1}{2 r^{2}}\right)=\tan ^{-1}\left(\frac{2}{4 r^{2}}\right)$
$=\tan ^{-1}\left(\frac{(2 r+1)-(2 r-1)}{1+(2 r+1)(2 r-1)}\right)=\tan ^{-1}(2 r+1)-\tan ^{-1}(2 r-1)$
Thus,
$\displaystyle\sum_{r=1}^{n} \tan ^{-1}\left(\frac{1}{2 r^{2}}\right)=\displaystyle\sum_{r=1}^{n}\left[\tan ^{-1}(2 r+1)-\tan ^{-1}(2 r-1)\right]$
$=\tan ^{-1}(2 n+1)=\tan ^{-1}(1)=\tan ^{-1}(2 n+1)-\pi / 4$
$\therefore \displaystyle \lim _{n \rightarrow \infty} \displaystyle\sum_{r=1}^{n} \tan ^{-1}\left(\frac{1}{2 r^{2}}\right)=\displaystyle\lim _{n \rightarrow \infty}\left[\tan ^{-1}(2 n+1)-\pi / 4\right]$
$=\tan ^{-1}(\infty)-\pi / 4=\pi / 2-\pi / 4=\pi / 4 .$