What is the limit of $\lim_{n\to \infty}\frac{1-2+3-4+\cdots+(-2n)}{\sqrt{n^2+1}}$? Find the $$\lim_{n\to \infty}\frac{1-2+3-4+\cdots+(-2n)}{\sqrt{n^2+1}}.$$
I thought to apply squeeze theorem so 
$$\frac{1-2+3-4+\cdots+(-2n)}{\sqrt{n^2+1}}\leq \frac{1+2+3+4+\cdots+(2n)}{\sqrt{n^2+1}}=\frac{n(2n+1)}{\sqrt{n^2+1}}$$ But I am unable to find a lower bound though. How can I do this?

You can do the following. For any $k$ we have $$1-2+3-4 + ... +(-2k) = (1-2) + (3-4) + \cdots +(2k-1 - 2k) =-k$$
