Evaluate lim$_{n\rightarrow\infty}$ $\frac{1-2+3-4+5-...............+\left(-2n\right)}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}$ 
Evaluate $$\lim_{n\rightarrow\infty}\frac{1-2+3-4+5-\cdots+\left(-2n\right)}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}.$$

My Approach 
Let $A_{n}=1+3+5+\cdots\left(2n-1\right)$ and $B_{n}=2+4+6+\cdots\left(2n\right)$, then $A_n=\frac{n}{2}\left(1+2n-1\right)=n^{2}$
and
$B_{n}=\frac{n}{2}\left(2+2n\right)=n+n^{2}$.
Therefore,
$$\lim_{n\rightarrow\infty}\frac{A_{n}-B_{n}}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}=\lim_{n\rightarrow\infty}\frac{-n}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}=-\frac{1}{2}.$$
But the book mentions that the answer is $0$.

One may write
$$
1-2+3-4+5-\cdots+\left(-2n\right)=\underbrace{-\left(2-1\right)}_{\color{red}{-1}}\:\underbrace{-\left(4-3\right)}_{\color{red}{-1}}-\cdots\underbrace{-\left(2n-(2n-1)\right)}_{\color{red}{-1}}=\color{red}{-n}
$$ giving
$$
\frac{1-2+3-4+5-\cdots+\left(-2n\right)}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}=\frac{-n}{n\sqrt{1+1/n^{2}}+n\sqrt{1-1/n^{2}}} \to -\frac12
$$ as $n\to \infty$.
Your result is correct to me.
