0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.