Stuck on Calculus 2 homework question about infinite series $\sum\frac1{n^2-3n+1}$ The question ask "Given the following infinite series determine if they are convergent or divergent (show your work on how you come to the decision)" the problem itself is: 
$$\sum\frac1{^2−3+1}$$
 A: use that $$\frac{1}{n^2-3n+1}\le \frac{2}{n^2}$$
A: Most of the time, you'll want to look at the "dominating" parts of the numerator and denominator to get an idea of whether it's convergent or not. The dominating part in the denominator is $n^2$ and in the numerator is $1$. Regular comparison works if you can find the smallest $n$ such that the regular comparison to $\frac{1}{n^2}$ is true, but that's slightly more tedious sometimes (in my opinion). So, let's use limit comparison with $\frac{1}{n^2}$ to see what happens (note that $\frac{1}{n^2}$ converges because it's p-series with $p=2$). So, $a_n$ is our original series function and let $b_n=\frac{1}{n^2}$. Then, $$\lim_{n\to\infty}\dfrac{\frac{1}{n^2}}{\frac{1}{n^2-3n+1}}=\lim_{n\to\infty}\dfrac{n^2-3n+1}{n^2}=\lim_{n\to\infty}\dfrac{1-\frac{3}{n}+\frac{1}{n^2}}{1}=1$$
which is finite and non-zero (To take the limit, I just multiplied top and bottom by $\frac{1}{n^2}$ for clarity). So, by the limit comparison test, both of these series have the same convergence behavior and hence both converge (since $\frac{1}{n^2}$ converges).
