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: 
Since the denominator has two linear factors, you can set up your partial fraction decomposition as

$$\frac{x}{(x+4)(x-1)} = \frac{A}{x+4} + \frac{B}{x-1}$$

This implies

$$ x = A(x-1) + B(x+4) $$

So $A+B=1$ and $4B-A=0$ $\Rightarrow$ $(A,B) = (\frac{4}{5},\frac{1}{5})$.

This means that

$$\int \frac{x}{(x-4)(x+1)} dx = \frac{1}{5} \int \frac{4}{x+4} + \frac{1}{x-1}dx $$

Then you can continue from here.

This is my first response, so I would very much appreciate any feedback on formatting or other site etiquette I may be unaware of. I hope this helps!
$$\int{\frac{xdx}{x^2 + 3x -4}}=\frac{1}{2}(\int{\frac{(2x+3)dx}{x^2 + 3x -4}}-3\int{\frac{dx}{x^2 + 3x -4}})=\frac{1}{2}(\int{\frac{(2x+3)dx}{x^2 + 3x -4}}+\frac{3}{5}(\int{\frac{dx}{x +4}}-\int{\frac{dx}{x -1}}))=\frac{1}{2}\ln{|x^2+3x-4|}+\frac{3}{5}\ln{|x+4|} -\frac{3}{5}\ln{|x-1|}+C=\frac{1}{2}\ln{|x+4|}+\frac{1}{2}\ln{|x-1|}+\frac{3}{10}\ln{|x+4|} -\frac{3}{10}\ln{|x-1|}+C=\frac{4}{5}\ln|x+4|+\frac{1}{5}\ln|x-1|+C$$