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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?

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

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Note that we can write

$$\begin{align} \sum_{k=1}^{2n}(-1)^{k-1}k&=\sum_{k=1}^{n}(2k-1)-\sum_{k=1}^n (2k)\\\\ &=\sum_{k=1}^n (-1)\\\\ &=-n \end{align}$$

Then, we have

$$\lim_{n\to \infty}\frac{1}{\sqrt{n^2+1}}\sum_{k=1}^{2n}(-1)^{k-1}k=-1$$

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