Question

For any positive integer $n$, let $S_{n}:(0, \infty) \rightarrow R$ be defined by
$S_{n}(x)=\displaystyle\sum_{k=1}^{n} \cot ^{-1}\left(\frac{1+k(k+1) x^{2}}{x}\right),$
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. $S _{10}( x )=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1+11 x ^{2}}{10 x }\right)$, for all $x >0$
• B. $\displaystyle\lim _{n \rightarrow \infty} \cot \left(S_{n}(x)\right)=x$, for all $x>0$
• C. The equation $S _{3}( x )=\frac{\pi}{4}$ has a root in $(0, \infty)$
• D. $\tan \left(S_{n}(x)\right) \leq \frac{1}{2}$, for all $n \geq 1$ and $x>0$

Answer 1

Answer: (A), (B)

$S_{n}=\displaystyle\sum_{k=1}^{n} \cot ^{-1}\left\{\frac{1+k(k+1) x^{2}}{x}\right\} $;
$ \displaystyle\sum_{k=1}^{n} \cot ^{-1}\left\{\frac{k(k+1) x^{2}+1}{(k+1) x-k x}\right\}$
$S_{n}=\displaystyle\sum_{k=1}^{n} \cot ^{-1}(k x)-\cot ^{-1}(k+1) x$
$t_{1}=\cot ^{-1}(x)-\cot ^{-1}(2 x)$
$t_{2}=\cot ^{-1}(2 x)-\cot ^{-1}(3 x)$
$t_{3}=\cot ^{-1}(3 x)-\cot ^{-1}(4 x)$
:
$t_{n}=\cot ^{-1}(n x)-\cot ^{-1}((n+1) x)$
$S_{n}=\cot ^{-1}(x)-\cot ^{-1}((n+1) x)$
$\Rightarrow S_{n}=\cot ^{-1}\left(\frac{(n+1) x^{2}+1}{n x}\right)$
$S_{10}=\cot ^{-1}\left(\frac{11 x^{2}+1}{10 x}\right)=\frac{\pi}{2}-\tan ^{-1}\left(\frac{11 x^{2}+1}{10 x}\right)$
(B) $\displaystyle\lim _{n \rightarrow \infty} \cot \left(S_{n}(x)\right)=\displaystyle\lim _{n \rightarrow \infty} \cot \left(\cot ^{-1}\left(\frac{(n+1) x^{2}+1}{n x}\right)\right)$
$=\displaystyle\lim _{n \rightarrow \infty} \frac{n x^{2}+x^{2}+1}{n x}=x$
(C) $S_{3}(x)=\cot ^{-1}\left(\frac{4 x^{2}+1}{3 x}\right)=\frac{\pi}{4} \Rightarrow \frac{1+4 x^{2}}{3 x}=1$
$\Rightarrow 4 x^{2}-3 x+1=0$ have imaginary roots
(D) $\tan \left(S_{n}(x)\right)=\tan \left(\cot ^{-1}\left(\frac{1+(n+1) x^{2}}{n x}\right)\right)$
$=\frac{n x}{1+(n+1) x^{2}}=\frac{1}{\frac{1}{n x}+\frac{(n+1) x}{n}}$