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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.

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    Compare with $$\sum_{k = 1}^{\infty} \frac{1}{k^2}.$$ – 2017-02-04
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    @DanielFischer How can I do this? – 2017-02-05
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    Show that $\dfrac{k^2 P(k)}{Q(k)}$ is bounded. Perhaps it helps you see it better to look at $\dfrac{k^{\deg Q - \deg P} P(k)}{Q(k)}$. – 2017-02-05
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    @DanielFischer Ok, so we proved that the sequence is bounded. What next to show that it is convergent? – 2017-02-05
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    Well, if you know that's bounded, you know $$\biggl\lvert \frac{P(k)}{Q(k)}\biggr\rvert \leqslant \frac{c}{k^2}$$ for some $c$. – 2017-02-05
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    @DanielFischer Ok, I understand this too. What next? I am new to this subject. – 2017-02-06
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    Compare with the **series** $\sum_k\frac{1}{k^2}$. See [comparison test](https://en.wikipedia.org/wiki/Direct_comparison_test). – 2017-02-06
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    @M.Stefan: Have you studied the concept of "series" and convergence tests for series? – 2017-02-07

1 Answers 1

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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 side of (3) is limited (recall that every convergent sequence is limited). 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 and, hence, convergent.