Question

Solve the following:
(a) ${cot}^{-1}(1+\frac{3}{4})+{cot}^{-1}(2^2+\frac{3}{4})+{cot}^{-1}(3^2+\frac{3}{4})+.....\infty$
(b) Find the sum of the series
${sin}^{-1}\left(\frac{1}{\sqrt{2}}\right)+{sin}^{-1}\left(\frac{\sqrt{2}-1}{\sqrt{6}}\right)+...+{sin}^{-1}\left[\frac{\sqrt{n}-\sqrt{n-1}}{\surd \left\{n\right(n+1\left)\right\}}\right]+....\infty$

Answer 1

(a) $T_n={cot}^{-1}(n^2+\frac{3}{4})={cot}^{-1}\frac{4n^2+3}{4}$
$={tan}^{-1}\frac{4}{4(n^2+\frac{3}{4})}={tan}^{-1}\frac{1}{1+(n^2-\frac{1}{4})}$
$={tan}^{-1}\frac{(n+\frac{1}{2})-(n-\frac{1}{2})}{1+(n+\frac{1}{2})(n-\frac{1}{2})}$
${tan}^{-1}(n+\frac{1}{2})-{tan}^{-1}(n-\frac{1}{2})$
putting $n=1,2,3,....,n$ and adding
$S_n={tan}^{-1}(n+\frac{1}{2})-{tan}^{-1}\frac{1}{2}$
$\therefore$ $S_{\infty }=\frac{\pi }{2}-{tan}^{-1}\frac{1}{2}={cot}^{-1}\frac{1}{2}={tan}^{-1}2$
(b) $T_n={sin}^{-1}[\frac{1}{\sqrt{n}}\sqrt{\left(\frac{n}{n+1}\right)}+\sqrt{\left(\frac{n-1}{n}\right)}\frac{1}{\sqrt{n+1}}]$
$={sin}^{-1}[\frac{1}{\sqrt{n}}\sqrt{(1-\frac{1}{n+1})}-\sqrt{(1-\frac{1}{n})}\frac{1}{\sqrt{n+1}}]$
If $sin\theta =\frac{1}{\sqrt{n}},cos\theta =\sqrt{(1-\frac{1}{n})}$
If $sin\varphi =\frac{1}{\sqrt{n+1}},cos\varphi =\sqrt{(1-\frac{1}{n+1})}$
L.H.S.$={sin}^{-1}(sin\theta cos\varphi -cos\theta sin\varphi )$
$={sin}^{-1}sin(\theta -\varphi )$
$=\theta -\varphi ={sin}^{-1}\frac{1}{\sqrt{n}}-{sin}^{-1}\frac{1}{\sqrt{n+1}}$
Now put $n=1,2,3,....$ and add.
$S_n={sin}^{-1}1-{sin}^{-1}\frac{1}{\sqrt{n+1}}$
$\therefore$ $S_{\infty }={sin}^{-1}1=\frac{\pi }{2}$