Question

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

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

Answer 1

Answer: (A), (B)

$\underset{n\to \infty }{lim}\cot \left(S_n\right(x\left)\right)=x,$ for all $x>0$

$S_n\left(x\right)=\sum _{k=1}^n{\cot }^{-1}\left(\frac{1+k(k+1)x^2}{x}\right)$
$=\sum _{k=1}^n{\tan }^{-1}\left(\frac{x}{1+\left(kx\right)\cdot \left(\right(k+1\left)x\right)}\right)$
$=\sum _{k=1}^n{\tan }^{-1}\left(\frac{(k+1)x-kx}{1+\left(kx\right)\cdot \left(\right(k+1\left)x\right)}\right)$
$=\sum _{k=1}^n{\tan }^{-1}(k+1)x-{\tan }^{-1}\left(kx\right)$
$\therefore S_n\left(x\right)={\tan }^{-1}(n+1)x-{\tan }^{-1}\left(x\right)$
$={\tan }^{-1}\left(\frac{nx}{1+(n+1)x^2}\right)$
For $n=10,$ we have
$S_{10}\left(x\right)={\tan }^{-1}\left(\frac{10x}{1+11x^2}\right)$
$\Rightarrow S_{10}\left(x\right)={\cot }^{-1}\left(\frac{1+11x^2}{10x}\right)$
$\therefore S_{10}\left(x\right)=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{1+11x^2}{10x}\right)$


$S_n\left(x\right)={\tan }^{-1}\left(\frac{nx}{1+(n+1)x^2}\right)$
$\Rightarrow \cot \left(S_n\right(x\left)\right)=\left(\frac{1+(n+1)x^2}{nx}\right)$
Now,
$\underset{n\to \infty }{lim}\cot \left(S_n\right(x\left)\right)=\underset{n\to \infty }{lim}\left(\frac{1+(n+1)x^2}{nx}\right)$
$\therefore \underset{n\to \infty }{lim}\cot \left(S_n\right(x\left)\right)=x$


$S_3\left(x\right)={\tan }^{-1}\left(\frac{3x}{1+4x^2}\right)=\frac{\pi }{4}$
$\Rightarrow \frac{3x}{1+4x^2}=1$
$\Rightarrow 4x^2-3x+1=0$
no real root $(\because D=-7<0)$


$S_n\left(x\right)={\tan }^{-1}\left(\frac{nx}{1+(n+1)x^2}\right)$
$\Rightarrow \tan \left(S_n\right(x\left)\right)=\left(\frac{nx}{1+(n+1)x^2}\right)$
For $x=1:$
$\tan \left(S_n\right(x\left)\right)=\left(\frac{n}{n+2}\right)\ge \frac{1}{2}\forall n\ge 2$