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.

  • $\begingroup$ 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 .... $\endgroup$ – 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.

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

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