Does the series $\sum_{n=2}^\infty {\frac{n+2}{n^3-2n^2+1}}$ converge?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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!