Is $\sum_{n=1}^{\infty} \frac{(-1)^n(3n^3+4n^2)}{4+2n^5}$ divergent, conditionally convergent, or absolutely convergent?

Question

The first thing I did was create a sequence ($A$) for what was inside of the sum, then I created another sequence ($B$) that is related to $A$.

$$A=\frac{(-1)^n(3n^3+4n^2)}{4+2n^5}$$ $$B=\frac{1}{n^2}$$

Using the two equations I first checked to see if it was absolutely convergent. This was done by comparing $|A|$ to $B$.

I know that $|A| > B$ for all of n, I also know that $n^{-2}$ is convergent. This means that $|A|$ might be convergent or divergent. I don't know how to proceade after this part. Any suggestions?

Answer 1

Let $B_n={1\over n^{3\over 2}}$, $lim_{n\rightarrow+\infty}{|A_n|\over B_n}=0$ implies that $|A_n|$ is convergent.

Answer 2

You started well. It turns out that$$\lim_{n\to\infty}\frac{\left\lvert\frac{(-1)^n(3n^3+4n^2)}{4+2n^5}\right\rvert}{\frac1{n^2}}=\frac32.$$So, since the series $\displaystyle\sum_{n=1}^\infty\frac1{n^2}$ converges, your series converges absolutely.

Answer 3

It is absolutely convergent, because a polynomial is asymptotically equivalent to its leading term, so $$\frac{(3n^3+4n^2)}{4+2n^5}\sim_\infty\frac{3n^3}{2n^5}=\frac 32\cdot\frac1{n^2};$$ and the latter converges.

We use here that if two series with positive terms have asymptotically equivalent general terms, they both converge or diverge.