If {x_n} = frac{{1 - 2 + 3 - 4 + 5 - 6 + dots | vedclass

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(B) The numerator is the sum of the series $1 - 2 + 3 - 4 + \dots - 2n$.
This can be grouped as $(1 - 2) + (3 - 4) + \dots + ((2n-1) - 2n)$.
There a

Vedclass · Accepted Answer

$-\frac{1}{3}$

Vedclass · Answer

(B) The numerator is the sum of the series $1 - 2 + 3 - 4 + \dots - 2n$.
This can be grouped as $(1 - 2) + (3 - 4) + \dots + ((2n-1) - 2n)$.
There are $n$ such pairs,each summing to $-1$.
Thus,the numerator is $n 	imes (-1) = -n$.
Now,$\mathop {\lim }\limits_{n 	o \infty } {x_n} = \mathop {\lim }\limits_{n 	o \infty } \frac{-n}{\sqrt {n^2 + 1} + \sqrt {4n^2 - 1}}$.
Dividing the numerator and denominator by $n$:
$\mathop {\lim }\limits_{n 	o \infty } \frac{-1}{\sqrt {1 + \frac{1}{n^2}} + \sqrt {4 - \frac{1}{n^2}}} = \frac{-1}{\sqrt{1+0} + \sqrt{4-0}} = \frac{-1}{1+2} = -\frac{1}{3}$.

If ${x_n} = \frac{{1 - 2 + 3 - 4 + 5 - 6 + \dots - 2n}}{{\sqrt {{n^2} + 1} + \sqrt {4{n^2} - 1} }},$ then $\mathop {\lim }\limits_{n \to \infty } {x_n}$ is equal to

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