Question

For any positive integer $n$, let $S_n:(0,\infty )\to R$ be defined by
$S_n(x)=\sum _{k=1}^n{\cot }^{-1}\left(\frac{1+k(k+1)x^2}{x}\right),$
where for any $x\in R,{\cot }^{-1}(x)\in (0,\pi )$ and ${\tan }^{-1}(x)\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$. Then which of the following statements is(are) TRUE?

• A. $S_{10}(x)=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{1+11x^2}{10x}\right)$, for all $x>0$
• B. $\underset{n\to \infty }{\lim }\cot \left(S_n(x)\right)=x$, for all $x>0$
• C. The equation $S_3(x)=\frac{\pi }{4}$ has a root in $(0,\infty )$
• D. $\tan \left(S_n(x)\right)\le \frac{1}{2}$, for all $n\ge 1$ and $x>0$

Answer 1

Answer: (A), (B)

$S_n=\sum _{k=1}^n{\cot }^{-1}\left\{\frac{1+k(k+1)x^2}{x}\right\}$;
$\sum _{k=1}^n{\cot }^{-1}\left\{\frac{k(k+1)x^2+1}{(k+1)x-kx}\right\}$
$S_n=\sum _{k=1}^n{\cot }^{-1}(kx)-{\cot }^{-1}(k+1)x$
$t_1={\cot }^{-1}(x)-{\cot }^{-1}(2x)$
$t_2={\cot }^{-1}(2x)-{\cot }^{-1}(3x)$
$t_3={\cot }^{-1}(3x)-{\cot }^{-1}(4x)$
:
$t_n={\cot }^{-1}(nx)-{\cot }^{-1}((n+1)x)$
$S_n={\cot }^{-1}(x)-{\cot }^{-1}((n+1)x)$
$\Rightarrow S_n={\cot }^{-1}\left(\frac{(n+1)x^2+1}{nx}\right)$
$S_{10}={\cot }^{-1}\left(\frac{11x^2+1}{10x}\right)=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{11x^2+1}{10x}\right)$
(B) $\underset{n\to \infty }{\lim }\cot \left(S_n(x)\right)=\underset{n\to \infty }{\lim }\cot \left({\cot }^{-1}\left(\frac{(n+1)x^2+1}{nx}\right)\right)$
$=\underset{n\to \infty }{\lim }\frac{n{x}^2+x^2+1}{nx}=x$
(C) $S_3(x)={\cot }^{-1}\left(\frac{4x^2+1}{3x}\right)=\frac{\pi }{4}\Rightarrow \frac{1+4x^2}{3x}=1$
$\Rightarrow 4x^2-3x+1=0$ have imaginary roots
(D) $\tan \left(S_n(x)\right)=\tan \left({\cot }^{-1}\left(\frac{1+(n+1)x^2}{nx}\right)\right)$
$=\frac{nx}{1+(n+1)x^2}=\frac{1}{\frac{1}{nx}+\frac{(n+1)x}{n}}$