The value of mathop {lim }limits_{n to infty | vedclass

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Question

(A) We know that the sum of the first $n$ natural numbers is given by $\sum n = \frac{n(n + 1)}{2}$.Substituting this into the limit expression:$\math

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(A) We know that the sum of the first $n$ natural numbers is given by $\sum n = \frac{n(n + 1)}{2}$.Substituting this into the limit expression:$\mathop {\lim }\limits_{n 	o \infty } \frac{1 - n^2}{\frac{n(n + 1)}{2}}$$= \mathop {\lim }\limits_{n 	o \infty } \frac{2(1 - n)(1 + n)}{n(n + 1)}$$= \mathop {\lim }\limits_{n 	o \infty } \frac{2(1 - n)}{n}$$= \mathop {\lim }\limits_{n 	o \infty } 2 \left( \frac{1}{n} - 1 ight)$As $n 	o \infty$,$\frac{1}{n} 	o 0$.Therefore,the limit is $2(0 - 1) = -2$.

The value of $\mathop {\lim }\limits_{n \to \infty } \frac{1 - n^2}{\sum n}$ is

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