Question

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

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. 1
• D. -1

Answer 1

The correct option is B

$-\frac{1}{2}$

${lim}_{n\to \infty }\left[\frac{1-2+3-4+5-6+\cdots (2n-1)-2n}{\sqrt{n^2+1}+\sqrt{n^2-1}}\right]$
$={lim}_{n\to \infty }\left[\frac{(1+3+5+\cdots 2n-1)-(2+4+6+\cdots 2n)}{(\sqrt{n^2+1}+\sqrt{n^2-1})}\right]$
$={lim}_{n\to \infty }\left[\frac{\frac{n}{2}(1+2n-1)-\frac{n}{2}(2+2n)}{(\sqrt{n^2+1}+\sqrt{n^2-1})}\right]$
$={lim}_{n\to \infty }\left[\frac{n^2-n(n+1)}{(\sqrt{n^2+1}+\sqrt{n^2-1})}\right]$
$={lim}_{n\to \infty }\left[\frac{-n}{(\sqrt{n^2+1}+\sqrt{n^2-1})}\right]$
Dividing the numerator and the denominator by n:
$={lim}_{n\to \infty }\left[\frac{-1}{\sqrt{1+\frac{1}{n^2}+\sqrt{1-\frac{1}{n^2}}}}\right]$
$=\frac{-1}{1+1}=\frac{-1}{2}$