The sum of the infinite series $\cot ^{-1}\left(\frac{7}{4}\right)+\cot ^{-1}\left(\frac{19}{4}\right)+\cot ^{-1}\left(\frac{39}{4}\right)+\cot ^{-1}\left(\frac{67}{4}\right)+\ldots$. is :
Solution
<p>$$\begin{aligned}
& \cot ^{-1}\left(\frac{7}{4}\right)+\cot ^{-1}\left(\frac{19}{4}\right)+\cot ^{-1}\left(\frac{39}{4}\right)+\cot ^{-1}\left(\frac{67}{4}\right)+\ldots \\
& T_r=\cot ^{-1}\left(\frac{4 r^2+3}{4}\right) \\
& T_r=\tan ^{-1}\left(\frac{1}{\left(\frac{3}{4}+r^2\right)}\right)
\end{aligned}$$</p>
<p>$$\begin{aligned}
& T_r=\tan ^{-1}\left(\frac{\left(r+\frac{1}{2}\right)-\left(r-\frac{1}{2}\right)}{1+r^2-1 / 4}\right) \\
& T_r=\tan ^{-1}\left(\frac{\left(r+\frac{1}{2}\right)-\left(r-\frac{1}{2}\right)}{1+\left(r+\frac{1}{2}\right)\left(r-\frac{1}{2}\right)}\right) \\
& T_r=\tan ^{-1}\left(r+\frac{1}{2}\right)-\tan ^{-1}\left(r-\frac{1}{2}\right) \\
& T_1=\tan ^{-1}\left(\frac{3}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right)
\end{aligned}$$</p>
<p>$$\begin{aligned}
& T_2=\tan ^{-1}\left(\frac{5}{2}\right)-\tan ^{-1}\left(\frac{3}{2}\right) \\
& \vdots \qquad\qquad \vdots \qquad \qquad\qquad \vdots \\
& T_n=\tan ^{-1}\left(\frac{2 n+1}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right) \\
& \Sigma T_r=\tan ^{-1}\left(\frac{2 n+1}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right) \\
& \Sigma T_r=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{2}\right)
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Inverse Trigonometric Functions · Topic: Domain and Range
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