Comparison test for $\sum\limits_{n=1}^\infty \frac{8^n}{7^n - n^2}$ Using the comparison test I'm trying to figure out the ratio and $$b_N$$ when
$$a_N=\sum_{n = 1}^\infty \frac{8^n}{7^n - 1n^2}$$
getting the highest terms would it be ?  
$$b_N=\left(\frac{8}{7}\right)^n$$
and because $$\frac{8}{7} > 1$$ the series diverges? Not sure if I'm on right course with $b_N$ or how to get the ratio. 
 A: You already have the right ideas for the solution, you just need to carefully put them together. Yes, your second series can be, and should be, the series of
$$\sum_{n = 1}^\infty b_n=\sum_{n = 1}^\infty\frac{8^n}{7^n}=\sum_{n = 1}^\infty\left(\frac{8}{7}\right)^n \qquad \text{i.e.} \qquad b_n=\frac{8^n}{7^n}=\left(\frac{8}{7}\right)^n.$$
Note that this is a geometric series with the common ratio $r=\frac{8}{7}$, and $|r|=\frac{8}{7}>1$, so this is a divergent geometric series.
Now, you need to show that you can make a useful comparison of the given series $\sum_{n = 1}^\infty a_n$ with this series $\sum_{n = 1}^\infty b_n$. There are two comparison tests: the so-called Direct or Basic Comparison Test and the Limit Comparison Test. Either one will work here.
To apply the Direct or Basic Comparison Test, you observe that
$$\frac{8^n}{7^n-n^2}>\frac{8^n}{7^n}, \qquad \text{i.e.} \qquad a_n>b_n,$$
and since $\sum_{n = 1}^\infty b_n$ diverges, the series $\sum_{n = 1}^\infty a_n$ of larger terms diverges as well.
To apply the Limit Comparison Test, you need to evaluate the limit
$$\lim_{n\to\infty} \frac{a_n}{b_n}=\lim_{n\to\infty} \frac{\frac{8^n}{7^n-n^2}}{\frac{8^n}{7^n}}=\cdots=1,$$
which tells us that the two series either both converge or both diverge. And since one of them is divergent, so is the other one.
A: You really answered the question without realizing it. You used more of a comparison test though(but you need the ratio test first).Think about it like this: We have a hunch that this will diverge (by your reasoning and work) so if we can COMPARE aN with a series that is smaller, but we know diverges then aN will also diverge.
Let take the series of bN = Sum from 1 to infinity of (8/7)n then we can see that aN is always larger(since we are dividing by a smaller number in every term). Now that we have our smaller series it remains to determine if that series, bN, diverges. We can use the ratio test and as you pointed out since (8/7)^n >1 for all n, then bN will diverge.
Therefore, by the comparison test aN diverges.
