Can I use the comparison test for the following problem? $$\sum _{n=1}^{\infty }\:\frac{1}{n^2-6n+10}$$
The denominator has a negative coefficient so i'm not sure if its valid to compare it to a p-series. Does this effect anything, or is it valid to to say this is convergent because it's less than 1/n^2?
Should I use the integral test instead?
For large $n$, the denominator is less than for $\frac{1}{n^2}$, which means the fraction is greater. Unfortunately, this comparison is the wrong direction for what you want. I recommend the limit comparison test, with the same series $(\sum \frac{1}{n^2})$.
Yes you can for every integer $n>3$ we have : $$\frac{1}{n^2-6n+10}=\frac1{(n-3)^2+1}\leq \frac{1}{(n-3)^2} $$
and we know that: $\sum_{n>3} \frac{1}{(n-3)^2} $ is convergent
Yes you can, $n^2 - 6n + 10 > \dfrac{n^2}{2}, n > 10$
You can do the limit comparison test. The limit comparison test is when one series "looks like" another series, but you don't want to go through all the little details of bounding precisely.
$$\lim_{n\to\infty}\frac{1}{n^2-6n+10}\frac{n^2}{1}=\lim_{n\to\infty}\frac{1}{1-\frac{6}{n}+\frac{10}{n^2}}=1$$
So they both converge.
By considering the logarithmic derivative of the Weierstrass product for the $\sinh $ function we have that: $$\sum_{n\geq 1}\frac{1}{(n-3)^2+1}=\frac{7}{10}+\sum_{n\geq 0}\frac{1}{n^2+1}=\frac{6}{5}+\frac{\pi\coth\pi}{2}.$$
