Question

Solve the following:
(a) ${cot}^{-1}(2^2+\frac{1}{2})+{cot}^{-1}(2^3+\frac{1}{2^2})+{cot}^{-1}(2^4+\frac{1}{2^3})+.....\infty$
(b) ${tan}^{-1}\frac{1}{3}+{tan}^{-1}\frac{2}{9}+...+{tan}^{-1}\frac{2^{n-1}}{1+2^{2n-1}}$

Answer 1

(a) $T_n={cot}^{-1}(2^{n+1}+\frac{1}{2^n})={cot}^{-1}\frac{2^{n+1}.2^n+1}{2^n}$
$={tan}^{-1}\frac{2^n(2-1)}{1+2^{n+1}.2^n}$
$={tan}^{-1}\frac{2^{n+1}-2^n}{1+2^{n+1}.2^n}$
$={tan}^{-1}{2}^{n+1}-{tan}^{-1}{2}^n$
Putting $n=1,2,3,....,n$ and adding,we get
$S_n={tan}^{-1}{2}^{n+1}-{tan}^{-1}2$
$\therefore$ $S_{\infty }={tan}^{-1}\infty -{tan}^{-1}2$
$\frac{\pi }{2}-{tan}^{-1}2={cot}^{-1}2={tan}^{-1}\frac{1}{2}$
(b) $T_n={tan}^{-1}\frac{2^{n-1}(2-1)}{1+2^n.2^{n-1}}={tan}^{-1}{2}^n-{tan}^{-1}{2}^{n-1}$
Now put $n=1,2,3,....,n$ and add,
$S_n={tan}^{-1}{2}^n-{tan}^{-1}1$.
$S_{\infty }=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$.