Is This a Complex Sequence? I am taking a look at the sequence $\{(-2)^\frac{1}{k}\}_{k=1}^\infty$. 
First, I have never really seen this type of sequence before... is this a valid sequence? I am reading online about complex sequences now, and I am not sure if this would be considered as one. 
If this is some sort of valid sequence (complex or not), is it possible to determine whether this sequence is convergent and monotonic?
 A: 
is it possible to determine whether this sequence is [...] monotonic?

To speak of a sequence being monotonic, there needs to be a notion of order. The complex numbers cannot be ordered, however. That is, for two complex numbers $z_1,z_2$, you can't say whether $z_1 < z_2$ or $z_1 > z_2$ under usual circumstances. Thus, the sequence can't be monotonic, since to be monotonic you have to be able to see something is less than or greater than another thing in the first place.
You can see a proof by counterexample of this fact in this video by Dr. Peyam. Basically, if you can say the complex numbers are ordered, then since $0 \ne i$, you should have $0<i$ or $i<0$, but both lead to contradictions, and thus no ordering.

Now back to your first question, which is a bit more nuanced to approach:

First, I have never really seen this type of sequence before... is this a valid sequence? 

This would absolutely be a sequence of complex numbers, if defined appropriately. The $k$-th member of the sequence is the $k$-th root of $-2$, but the problem is that there will be precisely $k$ such satisfactory numbers, so it needs a bit more specification. In that light, I feel this is too ill-defined as a sequence to properly give one.
For example, which element would you suggest $k=5$ corresponds to, of these: $\sqrt[5] 2 e^{ni\pi/5}$, where $n=1,-1,3,-3,$ or $5$? These are all numbers such that, raised to the fifth power, they are $-2$, but these are also distinct complex numbers. You can see this and approximations for their values on Wolfram:

A complex sequence formally is defined as a function for $f : I \to \Bbb C$ where $I \subseteq \Bbb Z \cap [a,b]$ for some $a,b \in \Bbb R$. (Or, in words: it is a function from an interval of integers to the complex numbers. That interval would be the indices of the members of the sequence.) Thus, you can't have multiple potential values for a member of a sequence: functions only give a single "output" regardless of the "input" given. 
You have to pick one and only one possible root in this case, but which one?
That's why I feel, in this light, the sequence is ill-defined.
In case the image above is suggestive that perhaps we can choose all members of the sequence to be real numbers somehow - and thus manage to have an ordering to prove monotonicity - some members of the sequence will be purely imaginary. For example, take $k=6$. Wolfram returns the following values:

Thus, no worries. Monotonicity is still not possible.


is it possible to determine whether this sequence is convergent[?]

Okay so let's suppose we adapt some sort of convention for the sequence's members and thus only have one possible value for each member of the sequence. Some will be guaranteed to be imaginary, so we have no monotonicity. 
Does it converge, however?
Luckily, yes, it does: to $0$. You can show that (regardless of the means in which we choose the members of the sequence), their magnitude eventually approaches $0$. (That is, the distance of the numbers in the sequence from the origin eventually approaches zero.) All of the complex numbers a fixed distance $r$ from the origin form a circle of radius $r$ centered there, so you can informally imagine this argument in the vein of shrinking a circle to have a "radius" of $0$. More formally, note that
$$(-2)^{1/k} = \sqrt[k] 2 \cdot e^{ni\pi/k}$$
for some integer $n$ that depends on our choice of root. We take the magnitude of both sides. Noting that all complex numbers of the form $e^{i \theta}$ for a real number $\theta$ fall on the unit circle (and thus have a magnitude of $1$), we see
$$|(-2)^{1/k}| = |\sqrt[k] 2 \cdot e^{ni\pi/k}| = |\sqrt[k] 2 |\cdot| e^{ni\pi/k}| = \sqrt[k] 2$$
Clearly, then, if $k \to \infty$, the sequence's members converge in magnitude to $0$, and thus have to converge to $0$ themselves.

In summary:


*

*The sequence is ill-defined as-is. There exist $k$ possible $k$-th roots of a number in the complex plane, so which of those do you pick?

*The complex numbers cannot be ordered, and not all members of the sequence are real numbers. Thus this definitely could be a complex sequence, if defined more precisely, but that also means monotonicity is not at all possible.

*If we define the sequence better, it can be shown to converge to $0$ since the magnitude of the terms (regardless of how we choose the roots) approaches zero.
