Question

If ${\tan }^{-1}\frac{4}{7}+{\tan }^{-1}\frac{4}{19}+{\tan }^{-1}\frac{4}{39}+{\tan }^{-1}\frac{4}{67}+\cdots \infty =\frac{\pi }{4}+{\cot }^{-1}k,$ where $k\in Z,$ then the value of ${}^{\prime }{k}^{\prime }$ is

Answer 1

Let,
$S=7+19+39+67+\cdots +T_n-S=-7-19-39-67-\cdots -T_{n}0=7+12+20+28+\cdots -T_n\Rightarrow T_n=7+12+20+28+\cdots \Rightarrow T_n=7+\frac{n-1}{2}[24+8(n-2\left)\right]\Rightarrow T_n=7+4(n-1)(n+1)\Rightarrow T_n=4n^2+3$

Therefore, the series can be written as,
$S_1=\sum _{n=1}^{\infty }{\tan }^{-1}\frac{4}{4n^2+3}$
$=\sum _{n=1}^{\infty }{\tan }^{-1}\frac{1}{n^2+\frac{3}{4}}$
$=\sum _{n=1}^{\infty }{\tan }^{-1}\frac{1}{1+(n^2-\frac{1}{4})}$
$=\sum _{n=1}^{\infty }{\tan }^{-1}\frac{1}{1+(n-\frac{1}{2})(n+\frac{1}{2})}$
$=\sum _{n=1}^{\infty }{\tan }^{-1}\frac{(n+\frac{1}{2})-(n-\frac{1}{2})}{1+(n-\frac{1}{2})(n+\frac{1}{2})}$
$=\sum _{n=1}^{\infty }{\tan }^{-1}(n+\frac{1}{2})-{\tan }^{-1}(n-\frac{1}{2})$
$=\underset{n\to \infty }{lim}{\tan }^{-1}(n+\frac{1}{2})-{\tan }^{-1}\left(\frac{1}{2}\right)$
$=\frac{\pi }{2}-{\tan }^{-1}\frac{1}{2}$

$\Rightarrow \frac{\pi }{2}-{\tan }^{-1}\frac{1}{2}=\frac{\pi }{4}+{\cot }^{-1}k$
$\Rightarrow {\cot }^{-1}k=\frac{\pi }{4}-{\tan }^{-1}\frac{1}{2}\Rightarrow {\cot }^{-1}k={\tan }^{-1}1-{\tan }^{-1}\frac{1}{2}\Rightarrow {\cot }^{-1}k={\tan }^{-1}\frac{1}{3}={\cot }^{-1}3\Rightarrow k=3$