Test of convergence of a sequence involving polynomials

Question

Let $P,Q \in \Bbb R[X]$ be two polynomials with real coefficients such that $\deg (P) \le \deg(Q)-2$ and $Q(n) \ne 0$ for every $n \in \Bbb N, n\gt 0$. Show that the sequence defined by $$a_n= \sum_{k=1}^n \frac {P(k)}{Q(k)}$$ is convergent. I tried to prove that the sequence is a Cauchy sequence, but I couldn't do the entire proof.

Answer 1

I will just write down what was discussed in the comments.

Let $n=\deg(P)$, $m=\deg(Q)$ and write $P(k)=a_0+a_1k+\cdots+a_nk^n$ and $Q(k)=b_0+b_1k+\cdots+b_mk^m$. Since $\deg (P) \le \deg(Q)-2$, we have $$\tag{1} m-n\geq 2 \Rightarrow \frac{k^n}{k^{m}}\leq\frac{1}{k^2}.$$ On the other hand,

$$\tag{2}\frac{P(k)}{Q(k)}=\frac{a_0+a_1k+\cdots+a_nk^n}{b_0+b_1k+\cdots+b_mk^m}=\frac{k^n\left(\frac{a_0}{k^n}+\frac{a_1}{k^{n-1}}+\cdots+\frac{a_{n-1}}{k}+a_n\right)}{k^m\left(\frac{b_0}{k^m}+\frac{b_1}{k^{m-1}}+\cdots+\frac{b_{m-1}}{k}+b_m\right)}.$$

Since $$\tag{3}\frac{\left(\frac{a_0}{k^n}+\frac{a_1}{k^{n-1}}+\cdots+\frac{a_{n-1}}{k}+a_n\right)}{\left(\frac{b_0}{k^m}+\frac{b_1}{k^{m-1}}+\cdots+\frac{b_{m-1}}{k}+b_m\right)}\longrightarrow \frac{a_n}{b_m}~~~~\text{as}~~~~k\to\infty,$$

the sequence in the left-hand side of (3) is bounded (recall that every convergent sequence is bounded). So there exists $c\in \mathbb{R}$, $c > 0$, such that

$$\tag{4}\left|\frac{\left(\frac{a_0}{k^n}+\frac{a_1}{k^{n-1}}+\cdots+\frac{a_{n-1}}{k}+a_n\right)}{\left(\frac{b_0}{k^m}+\frac{b_1}{k^{m-1}}+\cdots+\frac{b_{m-1}}{k}+b_m\right)}\right|\leq c.$$

Applying (4) to (2) and using (1) we get

$$\tag{5}\left|\frac{P(k)}{Q(k)}\right|=\left|\frac{k^n\left(\frac{a_0}{k^n}+\frac{a_1}{k^{n-1}}+\cdots+\frac{a_{n-1}}{k}+a_n\right)}{k^m\left(\frac{b_0}{k^m}+\frac{b_1}{k^{m-1}}+\cdots+\frac{b_{m-1}}{k}+b_m\right)}\right|\leq \frac{c}{k^2}.$$

It follows from (5) that

$$\sum_{k=1}^\infty \left|\frac{P(k)}{Q(k)}\right|\leq \sum_{k=1}^\infty\frac{1}{k^2}<+\infty$$

that is, the series $\sum_{k=1}^\infty \frac{P(k)}{Q(k)}$ is absolutely convergent, hence, convergent.