# Proof of the sequence $(a_n) =\frac{\alpha n^2+ \beta n+\gamma}{an^2+bn+c}$ converges?

I'm trying to do a assignment about sequences and got a few problems in this question:

Define $$(a_n) =\frac{\alpha n^2+ \beta n+\gamma}{an^2+bn+c}$$. Let $$a, b, c, \alpha, \beta, \gamma$$ be real constants such that $$a, \alpha \neq 0$$ and $$an^2 + bn + c \neq 0$$, $$\forall \in \mathbb{N}$$. Show that $$(a_n)$$ converges.

My problem with this question is I think it does converges to $$\frac{\alpha}{a}$$ and they are both different from zero, by creation. But how can I show that?

• Divide both numerator and denominator by $n^2$. The $\beta/n$ terms go to zero. – player3236 Sep 16 at 20:31

HINT

We have that

$$a_n =\frac{\alpha n^2+ \beta n+\gamma}{an^2+bn+c}=\frac{\alpha + \beta \frac1n+\gamma\frac1{n^2}}{a+b\frac 1n+c\frac1{n^2}}$$

which is a standard trick for this kind of limit for ratio of polynomials $$a_n=\frac{p(n)}{q(n)}$$.

Refer also to the related

Another way to go, using asymptotic analysis: a polynomial is asymptotically equivalent to its leading term, therefore $$\frac{\alpha n^2+ \beta n+\gamma}{an^2+bn+c}\sim_\infty\frac{\alpha n^2}{an^2}=\frac\alpha a.$$

Let us compute the difference between the generic term of your sequence and its limit, and show that it tends to $$0$$:

$$r_n=a_n-\frac{\alpha}{a} =\frac{\alpha n^2+ \beta n+\gamma}{an^2+bn+c}-\frac{\alpha}{a}$$

($$r_n$$ as $$n$$th residual).

$$=\frac{\alpha n^2+ \beta n+\gamma}{an^2+bn+c}-\frac{\alpha}{a}*\frac{n^2+\tfrac{b}{a}n+\tfrac{c}{a}}{n^2+\tfrac{b}{a}n+\tfrac{c}{a}}$$

$$=\frac{\alpha n^2+ \beta n+\gamma}{an^2+bn+c}-\frac{\alpha(n^2+\tfrac{b}{a}n+\tfrac{c}{a})}{an^2+bn+c}$$

$$=\frac{ (\beta-\tfrac{ \alpha b}{a})n+\gamma-\tfrac{\alpha c}{a}}{an^2+bn+c}\tag{1}$$

(because $$a \ne 0$$) which indeed converges to $$0$$ because the degree (at most $$1$$) of the numerator is less than the degree (2) of the denominator.

Edit: An interest of expression (1) is that you can assert, by studying the sign of the equivalent

$$r_n \approx \underbrace{\frac{ (\beta-\tfrac{ \alpha b}{a})}{a}}_C*\frac{1}{n}$$

that (unless $$C=0$$), the sequence $$(a_n)$$ converges asymptotically to its limit $$\frac{\alpha}{a}$$ while being larger than its limit or smaller than its limit, i.e., having an increasing or decreasing (long term) behavior:

If $$C>0$$, $$(a_n)$$ is an asymptotically decreasing sequence; if $$C<0$$, is an increasing sequence.

$$\frac{\alpha n^2+ \beta n+\gamma}{an^2+bn+c}-\frac\alpha a=\frac{(a\beta-b\alpha)n+(a\gamma-c\alpha)}{a(an^2+bn+c)}\to0$$

because the degree of the denominator exceeds that of the numerator.

Note that this also works when $$\alpha=0$$ (but not $$a=0$$).

• Little detail: the last denominator should be $a(an^2+bn+c)$. – Jean Marie Sep 16 at 21:36
• @JeanMarie: yes, thank you. – Yves Daoust Sep 16 at 21:43