Determine whether the series converges or diverges $\sum\limits_{n=1}^\infty\frac{n^3+4n}{n^4+200}$ I know intuitively that the series is similar to 1/n, therefore I'm gonna guess this series will diverge. Am I on the right track here?
$\sum\limits_{n=1}^\infty\frac{n^3+4n}{n^4+200} > \sum\limits_{n=1}^\infty\frac{n^3}{n^4+200}$
Now $\sum\limits_{n=1}^\infty\frac{n^3}{n^4+200} =$ something along the lines of 1/n minus something? And then use p-series comparison?
 A: Use that $$\frac{n^3+4n}{n^4+200}>\frac{1}{n}$$ if $$4n^2>200$$
A: From $$200 < n^4\quad \Rightarrow \quad n^4+200 < 2n^4\quad \text{for} \quad \forall n >3\in\mathbb{N}$$
Then
$$\frac{n^3 + 4n}{n^4 + 200} >\frac{n^3}{n^4+200}>\frac{1}{2n}$$
Hence $$\sum_{n\geq1} \frac{1}{2n} = \infty$$
Original series diverge by comparison test.
A: Note that for sufficiently large $n$, we have $n^4 + 200 < 2n^4$. So we have $\frac{n^3}{n^4+200} > \frac{n^3}{2n^4} = \frac{1}{2n}$ for large enough $n$, which proves divergence.
A: Instead of manipulating inequalities, it's often much shorter to use equivalence of functions (which is a very precise notion from Asymptotic Analysis, and not a vague and informal ‘similarity’). The main result we use is this:

Let $\sum_n u_n$ and $\sum_n v_n$ two series with positive terms, such that $u_n$ and $v_n$ are asymptotically equivalent.
Then  $\sum_n u_n$ and $\sum_n v_n$ both converge or both diverge.

For the given series, we use that a polynomial function is asymptotically equivalent to its leading term, so
$$\frac{n^3+4n}{n^4+200}\sim_{n\to\infty}\frac{n^3}{n^4}=\frac1n,\quad\text{which diverges.}$$
