Can an alternating series diverge to $\pm\infty$ Consider the series
$$ \sum_{n=1}^{+\infty}(-1)^n n!$$
The series cannot converge because it doesn't satisfy the vanishing condition:
$$\lim_{n \to +\infty}(-1)^nn! \neq 0$$
My question is: can the series diverge to $\pm\infty$? I have tried to answer the question by studying the sequence of partial sums
$$s_n = \sum_{k=1}^n(-1)^kk!$$
hoping to prove that the sign of $\{s_n\}$ is itself alternating, which would prove that the original series cannot diverge to $\pm \infty$. However, I don't know how to prove that
$$n! \geq |s_{n-1}|, \forall n$$
Am I on the right path? Is my conjecture true? I have tried googling an answer but I haven't found anything useful, only a bunch of warning about errouneus applications of the Leibniz test.
I would be very grateful I you could help me.
 A: In general, an alternating sum that diverges can diverge to either $\infty$ or $-\infty$. Take the sums
$$
1-1+2-1+3-1+4-1+5-1+\cdots
$$ and
$$
1-1+1-2+1-3+1-4+1-5+\cdots
$$
for examples of either.

Your sum, however, does not diverge to $\infty$ or $-\infty$ (i.e., it does not have a generalized limit). Since you already know that the sequence does not converge, it is sufficient to additionally show that the sequence of partial sums has an alternating sign*. This can easily be proven as it follows from the fact that, for every $n\geq 2$, we have $n! > \sum_{k=1}^{n-1} k!$**.
Note that it is not required to show that it is also unbounded. For example, the alternating sum
$$1-1+1-1+1-1+\cdots$$ does not have an unbounded sequence of partial sums, yet it still diverges.

* This is because if the series does not converge, then you know the sequence of partial sums can either diverge to $\infty$, to $-\infty$, or neither. If you show the sequence of partial sums has alternating sign, then it cannot diverge to $\infty$ (because it goes below $0$ infinitely many times) and it cannot diverge to $-\infty$ (because it goes above $0$ infinitely many times).
** This is very simple to show, since
$$
n! = n\cdot (n-1)! > \sum_{k=1}^n (n-1)! >\sum_{k=1}^n k! > \sum_{k=1}^{n-1}k!$$
A: Clearly $(s_{2n})_{n\in {\mathbb N}}$ increases while $(s_{2n+1})_{n\in {\mathbb N}}$ decreases. These prevents $s_n$ from converging to $+\infty$ or $-\infty$.
