Does the series $\sum_{n=2}^\infty {\frac{n+2}{n^3-2n^2+1}}$ converge? Do the following converge:
$\sum_{n=2}^\infty {\frac{n+2}{n^3-2n^2+1}}$
For this one I think the answer is no I just can't prove it. I split it up into partial fractions and got: $\frac{3n+1}{n^2-n-1}-\frac{3}{n-1}$
but after that I'm stumped :(
The second part: $\sum_{n=1}^\infty {\frac{1}{n}+\frac{(-1)^n}{n^2}}$
This one I think also diverges, I tried to use comparison test but didn't get far.
I think I am missing something obvious for both questions.
 A: Say $u_n=\frac{n+2}{n^3-2n^2+1}$ and $v_n=\frac{1}{n^2}$.
Then $$\lim_{n\to\infty} \frac{u_n}{v_n} =\lim_{n\to\infty} \frac{\frac{n+2}{n^3-2n^2+1}}{\frac{1}{n^2}} $$ 
$$=\lim_{n\to\infty} \frac{1+\frac{2}{n}}{1-\frac{2}{n}+\frac{1}{n^3}}=1>0$$
This limit is finite and $\sum v_n$ is convergent by p-series test.
Hence by comparison test, the given series converges.
A: For the second one (which should have been a separate question) note the following:

If $\sum a_n$ diverges and $\sum b_n$ converges, then $\sum (a_n+b_n)$ diverges.

This follows directly from the observation that $\sum (a_n+b_n) = \sum a_n + \sum b_n$
Let $a_n = \frac{1}{n}$ and $b_n = \frac{(-1)^n}{n^2}$.  Each of these sums should be familiar to you and can immediately give you your conclusion.
A: Here is a fairly simple approach.  Note that for all $n\ge2$ the following inequalities hold
$$n^3-2n^2+1\ge \frac18 n^3$$
and 
$$n+2\le 2n$$
Therefore, we have
$$\frac{n+2}{n^3-2n^2+1}\le\frac{2n}{\frac18 n^3}=\frac{16}{n^2}$$
Inasmuch as $\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}{6}$, we can write
$$\sum_{n=2}^{\infty}\frac{n+2}{n^3-2n^2+1}\le 16\sum_{n=2}^{\infty}\frac{1}{n^2}=\frac83 \pi^2-16$$
And we are done!
