What do I compare the series $\sum_{i=2}^n \frac{i^2+1}{i^3-1}$ to? I have tried everything I can think of. I know that the answer is divergent and that you can use a comparison test, I just don't know what to compare it to.
$$\sum_{i=2}^n \frac{i^2+1}{i^3-1}$$
I would love any help possible.
 A: Making the numerator smaller and the denominator larger makes the terms smaller. If it still diverges after this change, then the original series must also diverge.
With that in mind, subtract 1 from the numerators and add 1 to the denominators.
How to figure this out: You have a numerator where the dominating term is $i^2$, and a denominator where the dominating term is $i^3$. So it would be nice if we could make those the only terms.
This time it worked out in that making this change made the fractions smaller, and we wanted to show divergence. But what if the $+$ and the $-$ had been the other way? Then you note that $i^2-1\geq \frac12i^2$ and $i^3+1\leq 2i^3$. It still works more or less the same way, but the transformation isn't quite as simple.
A: Compare it to $1/i$, since $\lim_{i\to\infty}\frac{i(i^2+1)}{i^3-1}=\lim_{i\to\infty}\frac{1+1/i^2}{1-1/i^3}=1$.
A: Understanding asymptotic behaviours would make your life much easier in those series.
You have $i^2+1 \approx i^2$ and $i^3-1 \approx i^3$ when i is very large. 
Thus,  $\frac{i^2+1}{i^3-1} \approx \frac{i^2}{i^3}=\frac{1}{i}$ when i is very large. 
So you compare with $\frac{1}{i}$.
Note: if you had a bad approximation, there's no worries because comparison test won't work. (try comparing this to $\frac{1}{i^2}$)
A: If you perform the long division, you should get
$$\frac{i^2+1}{i^3-1}=\frac 1{i}+\frac 1{i^2}+\frac 1{i^4}+\frac 1{i^6}+\cdots >\frac 1{i} $$
