Comparing divergent series Let $n$ be a positive integer and $n \geq 2$ , is there a method to show $\sum_{i=1}^{\infty} 1$ < $\sum_{x=1}^{\infty} x^n$?
 A: Echoing the comments, I don't think there is a standard way to compare the sizes of divergent series, at least as far as I have ever learned. Here are some methods off of the top of my head:
1) Construct a new sequence
$$
r(n) = \frac{\sum_{x=1}^{k}x^n}{\sum_{i=1}^{k}1}
$$
which measures the ratio of the partial sums of your two sequences. If this tends to a nonzero real limit, then you might say that the two sequences are "about the same size". If it tends to 0, $-\infty$, or $\infty$ then we know that one is much larger than the other. There are of course issues of being well-defined if you apply this to two arbitrary sequences and all, but I leave that to you to figure out.
2) Construct a new series
$$
D = \sum_{i=1}^{\infty}\left(i^n-1\right)
$$
which measures the exact difference of the two series. If this equals a real constant, then you might say that the two sequences are "practically the same size". If it diverges, then one is larger than the other.
Does that give you something along the lines of what you are looking for?
EDIT:
Well, as for more methods, you will need to begin dreaming them up yourself. Following along the ideas of another commenter, you could investigate Landau notation if you are not familiar with it, and try to analyze the behavior of the sequences of partial sums. This is about the most general and powerful framework I can think of, but I also wanted to explicitly avoid it in case you are not ready to work with asymptotics.
