# How to prove when this series of polynomials is convergent and when it is divergent?

Let $$P$$ and $$Q$$ be polynomials of degree $$k$$ and $$m$$ and suppose $$Q(n)\ne0\forall n\in\mathbb N$$.

Prove that $$\sum_{n=1}^\infty\frac{P(n)}{Q(n)}$$ is convergent if $$m\ge k+2$$ and divergent if $$m\le k+1$$.

Can someone point me in the right direction? I have tried to transform into something telescoping and see if the terms cancel each other out but with no result.

Other than that, the only thing I can think of is trying induction but I feel as if there most likely is a much better way to prove it using other thereoms. Possibly the tail lemma.

• Before you start to prove anything you should (I suggest) ask yourself how big/small (roughly speaking) the $n$-th term is. It seems to me that it should be rather like $n^{k-m}$. Oh, maybe I should compare my series with .... – ancientmathematician Nov 17 '18 at 14:17

If $$m \ge k+2$$ and $$n$$ is large enough, then there exists $$c_1>0$$ such that

$$\left| \frac{Q(n)}{P(n)}\right| \ge c_1n^{m-k} \ge c_1n^2$$

that is $$\left| \frac{P(n)}{Q(n)}\right| \le \frac{c_1}{n^{2}}$$

Hence by comparison test, we have the result.

Also, if $$m \le k+1$$, WLOG, we assume that $$P$$ and $$Q$$ have positive leading coefficient. (Suppose not, we study its negative instead).

If $$n$$ is large enough, there exists $$c_2>0$$ such that $$\frac{P(n)}{Q(n)} \ge c_2n^{k-m} \ge \frac{c_2}{n}$$

Again, by comparison test, we have the result.

• Thank you, that helped a lot! – CruZ Nov 18 '18 at 8:53