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: 
Its wrong ! the step $$A(x+4)+B(x-1) = x$$ is not true instead you should have $$Ax(x+4)+B(x-1)=x$$ As you can see we would then have to take $A=0$ and the purpose of partial fractions is gone

I prefer $$\frac{x}{x^2-3x+4}=\frac{1}{5}(\frac{1}{x-1})+\frac{4}{5}(\frac{1}{x+4})$$

another way is to write $x=l(2x-3)+m$ (because derivative of $x^2-3x+4=2x-3$) then we have $$\int \frac{x}{x^2-3x+4}=\int \frac{2x-3}{2(x^2-3x+4)}+\int \frac{3}{2(x^2-3x+4)}$$ which is easy
The decomposition into partial fractions supposes these are *proper* rational fractions, where the degree of the numerator is less than the *degree of the irreducible factor* in the denominator. When the given fraction has only simple poles, as is the case here, this means the numerator is a constant, and actually, there exists a formula for a simple pole $\alpha$ of the rational function $\frac{P(x)}{Q(x)}$: the coefficient $A\_\alpha$ for the pole $\alpha$ is given by
$$A\_\alpha=\frac{P(\alpha)}{Q'(\alpha)},$$
which leads to the decomposition
$$\frac x{x^2+3x-4}=\frac x{(x-1)(x+4)}=\frac 15\frac1{x-1}+\frac 45\frac1{x+4}$$
and the integral
$$\frac15\bigl(\ln|x-1|+4\ln|x+4|\bigr)+C=\frac15\ln\bigl(|x-1|(x+4)^4\bigr)+C.$$