The value of limlimits_{n rightarrow infty} 6 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $T_r = 	an^{-1}\left(\frac{1}{r^2+3r+3}ight)$.We can rewrite the argument as $\frac{(r+2)-(r+1)}{1+(r+2)(r+1)}$.Thus,$T_r = 	an^{-1}(r+2)

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Let $T_r = 	an^{-1}\left(\frac{1}{r^2+3r+3}ight)$.We can rewrite the argument as $\frac{(r+2)-(r+1)}{1+(r+2)(r+1)}$.Thus,$T_r = 	an^{-1}(r+2) - 	an^{-1}(r+1)$.The sum $S_n = \sum_{r=1}^{n} T_r$ is a telescoping sum:$S_n = (	an^{-1}3 - 	an^{-1}2) + (	an^{-1}4 - 	an^{-1}3) + \dots + (	an^{-1}(n+2) - 	an^{-1}(n+1))$.$S_n = 	an^{-1}(n+2) - 	an^{-1}2$.Using the formula $	an^{-1}x - 	an^{-1}y = 	an^{-1}\left(\frac{x-y}{1+xy}ight)$:$S_n = 	an^{-1}\left(\frac{(n+2)-2}{1+(n+2)(2)}ight) = 	an^{-1}\left(\frac{n}{2n+5}ight)$.Now,we evaluate the limit:$\lim\limits_{n ightarrow \infty} 6 	an(S_n) = \lim\limits_{n ightarrow \infty} 6 	an\left(	an^{-1}\left(\frac{n}{2n+5}ight)ight)$.$= \lim\limits_{n ightarrow \infty} \frac{6n}{2n+5} = \frac{6}{2} = 3$.

The value of $\lim\limits_{n \rightarrow \infty} 6 \tan \left\{\sum\limits_{r=1}^{n} \tan ^{-1}\left(\frac{1}{r^{2}+3 r+3}\right)\right\}$ is equal to

Similar Questions

Let $[P]$ denote the greatest integer $\leq P$. If $0 \leq a \leq 2$,then the number of integral values of $a$ such that $\lim _{x \rightarrow a}([x^2]-[x]^2)$ does not exist is:

The value of $\mathop {\lim }\limits_{n \to \infty } \frac{{{x^n}}}{{{x^n} + 1}}$ where $x < -1$ is

$\mathop {\lim }\limits_{x \to \frac{\pi^+}{2}} e^{[\cot x]}$ is equal to :-
(where $[.]$ is the greatest integer function)

If $0 < p < q$,then $\lim _{n \rightarrow \infty}\left(q^n+p^n\right)^{1 / n}$ is equal to :

The value of $\lim_{x \to 3} \frac{[x] - 3}{x - 3}$, where $[\cdot]$ denotes the greatest integer function, is...

Vedclass Test Series

Exam Paper Generator

Online Exam Module