The statement of l'Hopital's rule found in Rudin's Principles of Mathematical Analysis (page 109) is:
$5.13$ $\,\,$ Theorem $\,\,\,\,\,\,\,$ Suppose $f$ and $g$ are real and differentiable in $(a,b)$, and $g'(x)\neq 0$ for all $x\in (a,b)$, where $-\infty\leq a < b\leq +\infty$. Suppose
$$
\frac{f'(x)}{g'(x)}\to A\,\textrm{ as }\,x\to a.
$$
If
$f(x)\to 0$ and $g(x)\to 0$ as $x\to a$, or if $g(x)\to +\infty$ as $x\to a$, then
$$
\frac{f(x)}{g(x)}\to A\,\textrm{ as }\,x\to a.
$$
The analogous statement is of course also true if $x\to b$, or if $g(x)\to-\infty$ [...]. Let us note that we now use the limit concept in the extended sense of Definition $4.33$.
It seems that what $A$ is supposed to be is a little ambiguous. Real? Possibly infinite? However, we can resolve this enigma with a quick look at definition $4.33$:
$4.33$ $\,\,$ Definition $\,\,\,\,\,\,\,$ Let $f$ be a real function defined on $E\subset \Bbb{R}$. We say that
$$
f(t)\to A\,\textrm{ as }\,t\to x,
$$
where $A$ and $x$ are in the extended real number system, if for every neighborhood $U$ of $A$ there is a neighborhood $V$ of $x$ such that $V\cap E$ is not empty, and such that $f(t)\in U$ for all $t\in V\cap E$, $t\neq x$.
So indeed, $A$ is allowed to be infinite (it's in the extended real number system)! We also see that Rudin treats the cases of $A = \pm\infty$ in the proof of l'Hopital's rule, so if the limit of the quotient of derivatives is infinite, the original limit must have been also. That is, there is no example (let alone a simple one) where the limit of the quotient of derivatives is infinite, but the original limit is finite.
I can derive the sum quite easily, using partial fractions and the residue theorem.
Note that $$\frac{k^2}{k^3+1} = \frac13 \left [\frac{1}{k+1} + \frac{2 k-1}{k^2-k+1}\right ]$$
Now, note that
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k+1} = 1-\log{2}$$
For the other piece, note the fortuitous coincidence that
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} (2 k-1)}{k^2-k+1} = -\frac12 \sum_{k=-\infty}^{\infty} \frac{(-1)^{k} (2 k-1)}{k^2-k+1}$$
(One may see this by showing that the map $k \mapsto -k$ produces the $k+1$th term of the summand.)
Now, the sum on the RHS may be evaluated using residue theory. In general, one may show that
$$\sum_{k=-\infty}^{\infty} (-1)^k f(k) = -\pi \sum_n \operatorname*{Res}_{z=z_n} [\csc{\pi z} \, f(z)]$$
where the $z_n$ are the non-integral poles of $f$. In this case, the poles are at $z=e^{\pm i \pi/3}$, and we are also lucky to have an $f$ of the form $g'/g$, so that the residue at each pole is simply $1$. Therefore, we have
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} (2 k-1)}{k^2-k+1} = \frac{\pi}{2} \left ( \csc{\left (\pi e^{i \pi/3}\right )}+ \csc{\left (\pi e^{-i \pi/3}\right )} \right ) = \pi \, \text{sech}{\left ( \frac{\sqrt{3} \pi}{2}\right )}$$
Therefore,
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} k^2}{k^3+1} = \frac13 \cdot \left [ 1-\log{2} + \pi \, \text{sech}{\left ( \frac{\sqrt{3} \pi}{2}\right )}\right ]$$
Best Answer
It is possible to apply the rule, by brutal force (not recommended). $$\lim_{x\to \infty}[x^2-\sqrt{x^4+4}]=\lim_{x\to \infty} {1-\sqrt{1+4x^{-4}}\over x^{-2}}\\ \underset{H}{=}-\lim_{x\to\infty}{{{16x^{-5}\over2\sqrt{1+4x^{-4}}}}\over 2x^{-3}}=-\lim_{x\to\infty}{4x^{-2}\over \sqrt{1+4x^{-4}}}=0$$
Remark The calculations simplify substantially if we subsitute $y=2x^{-2}$ in the second expression. In this way we get $$2\lim_{y\to 0^+}{1-\sqrt{1+y^2}\over y}\underset{H}{=}2 \lim_{y\to 0^+}{-y\over\sqrt{1+y^2}}=0$$