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Question

Mathematics Question on Sequence and series

For any positive integer nn, let Sn:(0,)RS_{n}:(0, \infty) \rightarrow R be defined by
Sn(x)=k=1ncot1(1+k(k+1)x2x),S_{n}(x)=\displaystyle\sum_{k=1}^{n} \cot ^{-1}\left(\frac{1+k(k+1) x^{2}}{x}\right),
where for any xR,cot1(x)(0,π)x \in R, \cot ^{-1}(x) \in(0, \pi) and tan1(x)(π2,π2)\tan ^{-1}(x) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right). Then which of the following statements is(are) TRUE?

A

S10(x)=π2tan1(1+11x210x)S _{10}( x )=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1+11 x ^{2}}{10 x }\right), for all x>0x >0

B

limncot(Sn(x))=x\displaystyle\lim _{n \rightarrow \infty} \cot \left(S_{n}(x)\right)=x, for all x>0x>0

C

The equation S3(x)=π4S _{3}( x )=\frac{\pi}{4} has a root in (0,)(0, \infty)

D

tan(Sn(x))12\tan \left(S_{n}(x)\right) \leq \frac{1}{2}, for all n1n \geq 1 and x>0x>0

Answer

S10(x)=π2tan1(1+11x210x)S _{10}( x )=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1+11 x ^{2}}{10 x }\right), for all x>0x >0

Explanation

Solution

(A) S10(x)=π2tan1(1+11x210x)S _{10}( x )=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1+11 x ^{2}}{10 x }\right), for all x>0x >0
(B) limncot(Sn(x))=x\displaystyle\lim _{n \rightarrow \infty} \cot \left(S_{n}(x)\right)=x, for all x>0x>0