Question

$\underset{n\to \infty }{\lim }\frac{1-2+3-4+5-6+....+\left(2n-1\right)-2n}{\sqrt{n^2+1}+\sqrt{n^2-1}}$ is equal to

(a) $\frac{1}{2}$

(b) $-\frac{1}{2}$

(c) 1

(d) −1

Answer 1

(b) $-\frac{1}{2}$

$$\begin{gathered} \underset{n\to \infty }{\lim }\left[\frac{1-2+3-4+5-6+...\left(2n-1\right)-2n}{\sqrt{n^2+1}+\sqrt{n^2-1}}\right] \\ =\underset{n\to \infty }{\lim }\left[\frac{\left(1+3+5+...2n-1\right)-\left(2+4+6+...2n\right)}{\left(\sqrt{n^2+1}+\sqrt{n^2-1}\right)}\right] \\ =\underset{n\to \infty }{\lim }\left[\frac{{\displaystyle \frac{n}{2}}\left(1+2n-1\right)-{\displaystyle \frac{n}{2}}\left(2+2n\right)}{\left(\sqrt{n^2+1}+\sqrt{n^2-1}\right)}\right] \\ =\underset{n\to \infty }{\lim }\left[\frac{n^2-n\left(n+1\right)}{\left(\sqrt{n^2+1}+\sqrt{n^2-1}\right)}\right] \\ =\underset{n\to \infty }{\lim }\left[\frac{-n}{\left(\sqrt{n^2+1}+\sqrt{n^2-1}\right)}\right] \end{gathered}$$

Dividing the numerator and the denominator by n:

$$\begin{gathered} =\underset{n\to \infty }{\lim }\left[\frac{-1}{\sqrt{1+{\displaystyle \frac{1}{n^2}}}+\sqrt{1-{\displaystyle \frac{1}{n^2}}}}\right] \\ =\frac{-1}{1+1} \\ =\frac{-1}{2} \end{gathered}$$