The value of lim_{n to infty} [frac{1}{1 - n^ | vedclass

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

Question

(D) The given expression is $S_n = \sum_{k=1}^{n} \frac{k}{1 - n^2}$.Since the denominator $(1 - n^2)$ is independent of $k$, we can write $S_n = \fra

Vedclass · Accepted Answer

$\frac{-1}{8}$

Vedclass · Answer

(D) The given expression is $S_n = \sum_{k=1}^{n} \frac{k}{1 - n^2}$.Since the denominator $(1 - n^2)$ is independent of $k$, we can write $S_n = \frac{1}{1 - n^2} \sum_{k=1}^{n} k$.Using the formula for the sum of the first $n$ natural numbers, $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$.Thus, $S_n = \frac{n(n+1)}{2(1 - n^2)}$.We can factor the denominator: $1 - n^2 = -(n^2 - 1) = -(n-1)(n+1)$.So, $S_n = \frac{n(n+1)}{-2(n-1)(n+1)} = \frac{n}{-2(n-1)}$.Now, take the limit as $n 	o \infty$:$\lim_{n 	o \infty} \frac{n}{-2n + 2} = \lim_{n 	o \infty} \frac{n}{n(-2 + \frac{2}{n})} = \frac{1}{-2 + 0} = -\frac{1}{2}$.Note: The provided options do not contain the correct value $-\frac{1}{2}$. However, if the expression was $\sum_{k=1}^{n} \frac{k}{n^2}$, the limit would be $\frac{1}{2}$. Given the standard structure, the result is $-\frac{1}{2}$.

The value of $\lim_{n \to \infty} [\frac{1}{1 - n^2} + \frac{2}{1 - n^2} + \dots + \frac{n}{1 - n^2}]$ is

Similar Questions

$\lim _{x \rightarrow 1} (1 + \log _{e} x)^{1 / \log _{e} x}$ is equal to

$\mathop {\lim }\limits_{x \to \infty } \frac{{{{(2x + 1)}^{40}}{{(4x - 1)}^5}}}{{{{(2x + 3)}^{45}}}} = $

$\lim _{x \rightarrow \infty}\left[\sqrt{x^2+a x+b}-x\right]$ where $a < 0 < b$.

$\lim _{x \rightarrow 0} \frac{\left(2^x-1\right)(1+\sin x)^{\frac{2}{\sin x}}}{\log (1+2 x)} = $

If $\lim _{n \rightarrow \infty} \frac{1-(10)^n}{1+(10)^{n+1}}=\frac{-\alpha}{10}$,then $\alpha$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module