$ \lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}| $ Did I do it right? How to do the last step? I need to find:
$$ \lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}| $$
So I calculate:
$$ \lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}|  = \lim_{n \to +\infty} |\frac{\sum_{k=1}^{2n} (-1)^k \sqrt{k}}{\sqrt{n}} | $$
Denominator grows to $\infty$, so I can use stolz theorem:
$$\lim_{n \to +\infty} |\frac{\sum_{k=1}^{2n} (-1)^k \sqrt{k}}{\sqrt{n}} |  = \lim_{n \to +\infty} |\frac{\sqrt{2n}}{\sqrt{n} - \sqrt{n-1}} | \implies[\frac{\sqrt{2}}{0} ] $$
So I have to transform that a bit:
$$\lim_{n \to +\infty} |\frac{\sqrt{2n}(\sqrt{n} + \sqrt{n-1})}{n - n + 1} | = \lim_{n \to +\infty} |\sqrt{2}n + \sqrt{2n^2-2n}| \implies \infty - \infty$$
Another transformation:
$$\lim_{n \to +\infty} |\frac{2n^2 - 2n^2 + 2n}{\sqrt{2}n - \sqrt{2n^2-2n}}| = \lim_{n \to +\infty} |\frac{2n}{n(\sqrt{2} - \sqrt{2-\frac{2}{n}})}| = \lim_{n \to +\infty} |\frac{2}{\sqrt{2} - \sqrt{2-\frac{2}{n}}}| \implies[\frac{2}{0} ]$$
... and I feel like it doesn't lead nowhere. Can somebody tell what do I do wrong?

$$\begin{split}
\sum_{k=1}^{2n} (-1)^k \sqrt{k} &= \sum_{k=1}^{n} \left((-1)^{2k-1} \sqrt{2k-1} + (-1)^{2k} \sqrt{2k}\right)\\
&= \sum_{k=1}^{n} \left( \sqrt{2k}-\sqrt{2k-1}\right)\\
&= \sum_{k=1}^{n} \frac{1}{\sqrt{2k}+\sqrt{2k-1}}
\end{split}$$
Thus $$\sum_{k=1}^n\frac{1}{2\sqrt{2k}}\leq \sum_{k=1}^{2n} (-1)^k \sqrt{k}\leq \sum_{k=1}^n\frac{1}{2\sqrt{2k-1}}$$
We can use integrals to estimate the bounds
$$\frac 1 {2\sqrt 2}\int_1^{n+1}\frac{dx}{\sqrt x}\leq \sum_{k=1}^{2n} (-1)^k \sqrt{k}\leq \frac 1 2+ \int_2^{n+1}\frac{dx}{2\sqrt{2x-1}}$$
to see that $$\sum_{k=1}^{2n} (-1)^k \sqrt{k}\sim \sqrt{\frac n {2}}$$
Thus
$$\lim_{n \to +\infty} \frac{1}{\sqrt{n}} | \sum_{k=1}^{2n} (-1)^k \sqrt{k}|=\frac 1 {\sqrt 2}$$
