Question: 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?

Answer: 
**HINT**

Here it is an alternative way to solve it:

\begin{align\*}
\frac{x}{(x-1)(x+4)} & = \frac{(x-1) + 1}{(x-1)(x+4)}\\\\
& = \frac{1}{x+4} + \frac{1}{(x-1)(x+4)}\\\\
& = \frac{1}{x+4} + \frac{1}{5}\times\frac{(x+4) - (x-1)}{(x-1)(x+4)}\\\\
& = \frac{1}{x+4} + \frac{1}{5}\times\left[\frac{1}{x-1} - \frac{1}{x+4}\right]\\\\
& = \frac{1}{5(x-1)} + \frac{4}{5(x+4)}
\end{align\*}

Can you take it from here?
$$I=\int\frac{x}{x^2+3x+4}dx=\frac12\int\frac{2x+3}{x^2+3x+4}-\frac{3}{x^2+3x+4}$$
$$I=\frac12\ln|x^2+3x+4|-\frac32\int\frac{1}{(x+3/2)^2+7/4}dx$$
then you can let $u=x+3/2,\,du=dx$ so you have $u^2+(\sqrt{7}/2)^2=\frac74\left((2u/\sqrt{7})^2+1\right)$ now let $v=2u/\sqrt{7},\,dv=2/\sqrt{7}du$ now you have an integral of the form:
$$\int\frac{1}{v^2+1}dv$$ which can be easily solved using the substitution $v=\tan(t)$